60 JEE MAIN CHAPTERWISE EXPLORER
y = 2 – x 3 3
L(1, 1)
y = x – 2, x > 2 M(3, 1) ò ò 41. (c) : 1 - x2 dx = (1 - x)(1 + x) dx
-2
-2 – x 2 = 0
\ x = ± 1
Putting 1
Points –2, –1, 1, 3
(0, 0) A(1, 0) B(2, 0) C(3, 0) ìï 1 - x2 if | x |< 1
x = 1, x = 2, x = 3 if x < -1 and x ³ 1
\ |1 – x2 | = í - (1 - x 2 ))
îï(1
1 1
= [AL × AB + BC × CM] = [1 × 1 + 1 × 1] = 1 3
2 2 ò \ (1 - x2 ) dx
e x -2
38. (c) : As f (x) = 1 + ex
-1 1 3
e a e -a
\ f (a) = 1 + ea and f (–a) = 1 + e-a ò ò ò = (x2 - 1)dx + (1 - x2 )dx + (x2 - 1) dx
\ f (–a) + f (a) = 1
-2 -1 1
= 4 + 2 èæç 2 öø÷ + 20 = 28
3 3 3 3
f (a )
ò Now x g {x(1 - x)} dx = ò 42. (d) : 1 dx where a = b = 1
f (-a ) a cos x - b sin x
f (a ) let a = r cos q = 1
ò (1 - x)g {(1 - x)(x)} dx b = r sin q = 1
f (-a ) \ r = 2
q = tan –1( b/a)
b b
ò òusing f (x)dx = f (a + b - x) dx ò 1 1 1 sin x dx
cos x -
a a = 1
2
f (a ) f (a )
Þ 2 ò x g {x(1 - x)} dx = ò g {(1 - x) x} dx 2 2
f (-a ) f (-a ) 1 1 dx
2 cos(x + p/4)
Þ 2I 1 = I 2 ò =
\ I 2 = 2 ò = 1 p 1 dx
I1 1 2 sin çæè 2 pö
+ x + 4 ÷ø
p
p 2
ò ò 39. (b) : x f (sin x) dx = A f (sin x) dx ò = 1 x 1 dx
2 æ 2 3p ö cos çæè x 3p ö
0 0 2 sin çè + 8 ÷ø 2 + 8 ÷ø
p
pp
ò ò ò2 p
=
or A f (sin x) dx
2 f (sin x)dx = xf (sin x) dx sec2 çèæ 3 p + x ö
8 2 ÷ø
0 0 0 ò 1 dx
pp = tan çèæ x 3p ö
2 2 2 8 ø÷
2 òp 2 +
ò Þ A f (sin x) dx = ´ 2 f (sin x) dx 1 ´ 2 log x 3 p
2 2 2 2 8
0
0
pp = tan æçè + ö + c
÷ø
2 2
ò òÞ A f (sin x) = p f (sin x) dx
0 0 1 log tan èæç x 3 p ö
2 2 8 ø÷
Þ A = p = + + c
p/2 (sin x + cos x ) 2 p/2 sin x dx = Ax + B log sin (x – a) + C
sin(x - a)
(a) : dx = (sin x + cos x) dx
ò ò 40. ò 43. (b) :
0 (sin x + cos x) 2 0
p Þ Differentiating w.r.t. x both sides
= æ cos x + sin x ö 2 sin x B cos (x - a)
çè -1 ÷ø0
Þ sin(x - a ) = A + sin (x - a)
= 1 – (–1) = 2 Þ sin x = A sin (x – a) + B cos (x – a)
Integral Calculus 61
sin x = A (sin x cosa – cos x sina) p/2
+ B (cos x cos a + sin x sin a)
\ 2 ò sin3 q cos 2n + 1 q d q
sin x = sin x (A cos a + B sin a)
+ cos x(B cos a – A sin a) 0
Now solving A cos a + B sin a = 1 2[2 ´ (2n)(2n - 2)(2n - 4) ...4.2]
and B cos a – A sin a = 0 =
(A, B) = (cos a, sin a) (2n + 4)(2n + 2)(2n)(2n - 2)... 4.2
r = n 1 r 2 ´ 2 ´1
n =
(2n + 4)(2n + 2)
å 44. e 1
(b) : Lt r = 1 n =
n ®¥ (n + 2)(n + 1)
= 1 - 1 (by partial fraction)
1
n + 1 n + 2
ò = e x dx = e – 1
0
ò 4 3 esin x 3 dx b
45. (c): Given 1 x = F(k) – F(1) ò 48. (a), (c) : Let I = x f (x) dx
a
4 3 x 2 esin x 3 dx b
Þ ò x3 = F(k) – F(1) ò I = (a + b - x) f (a + b – x) dx
1 a
b b
Þ 64 e sin z dz = F(k) – F(1) where (x3 = z)
ò ò ò I = (a + b) f (a + b – x) dx – x f (a + b - x) dx
z a a
1
Þ [ F (z ) ]16 4 = F(k) – F(1) b b
Þ F(64) – F(1) = F(k) – F(1) ò òI = (a + b) f (x)dx - xf (x) dx
a a
Þ k = 64 a + b b a + b b
2 2
ò ò \ f f (a b x) dx
46. (b) : Lt ( tan t )0 x 2 I = (x) dx = + -
x ® 0 x sin x a a
t
tan x 2 ò 49. (a) : From given F(t) = f (t - y) g( y) dy
= Lt x sin x 0
t
x ® 0
ò = et - y y dy (By replacing y ® t – y in f (y))
tan x 2
= Lt 0
sin x 0 t
x ® 0 x 2
ò ò F(t) = - (t - q)eqdq = (t - q) eqdq
x
= Lt tan x 2 1 t 0
Lt sin x = 1 × 1 = 1
x ® 0 x 2 = (t eq )0 t – [(q – 1) eq] 0 t
= t(e t – 1) – (t – 1)e t – 1
x ® 0 x
= e t( t – t + 1) – t – 1
1 = e t – (t + 1)
47. (b) : ò x(1 - x) n dx 50. (b) : As f (x) = f ¢ (x) and f (0) = 1
0 f ¢ (x )
Þ f (x) = 1
Putting x = sin2 q
Þ log(f (x)) = x
dx = 2 sin q cos q dq Þ f (x) = e x + k
Þ f (x) = e x as f (0) = 1
and x = 0, q = 0 Now g(x) = x 2 – e x
x = 1, q = p/2
1 p/2
\ ò x(1 - x) n dx = ò sin2 q cos 2 n q (2 sin q cos q)dq
0 p
0
2 1 1
= 2 ò sin3 q cos 2n + 1q d q ò ò \ f (x) g (x) dx = ex (x2 - ex ) dx
0 0 0
é p/2 1 1
ê Using ò sin2n + 1 q cos 2n + 1q dq ò ò = x2e xdx - e2 x dx
êë 0 0 0 e 2 x ö1
2 ÷÷ø0
= [(2n)(2n - 2)...2][(2n)(2n - 2)...2] ù = [(x 2 – 2x + 2)e x] 01 – æ
çç
(4n + 2)(4n)(4n - 2)....2 ú è
û
62 JEE MAIN CHAPTERWISE EXPLORER
= (e – 2) – æ e2 - 1 ö e2 - 3 = 4p × æ p - 0 ÷öø = p 2
ççè 2 ÷÷ø = e – 2 2 çè 4
Using f n (x)e x dx = e x [f n (x) – f 1n (x) + f 2n ( x) + ...+ (–1) n f n( x)] 2
where f 1 , f 2, .... f n are derivatives of first, second ...nt h order. ò 55. (d) : Given f (x) dx = 3/4
0
51. (b) : Required area
0 1 2
ò ò ò = (3 + x) - (-x +1) + (3 - x) - (-x +1) dx + (3 - x) - (x -1) dx
1
-1 0
0 1 2
ò ò ò = 2(1 + x) dx + 2 dx + (4 - 2x) dx (0, 0) (2, 0)
-1 0 1 2 2 2
= 4 sq. units \ ò x f ¢(x) dx = xò f ¢(x) dx – ò f (x) dx
0 0 0
52. (a) : Lt 1p + 2 p + ..... + n p 1 = (x × f (x))02 - 3/4 = 2f(2) – 3
´ 4
n ® ¥ n p n
1 é æ 1 ö p æ 2 ö p æ n ö p ù 3
n ê çè n ÷ø çè n ÷ø çè n ÷ø ú = 0 – [\ f(2) = 0, curve having intercept
Lt + + .... + 4
= êë ûú
n ® ¥ = –3/4 2 units on xaxis.]
r=n 1 æ r ö p 1 x p dx 1 10 p
r = 1 n çè n ÷ø = 0 p + 1
å ò = ò 56. (d) : sin x dx
Lt =
n ® ¥
53. (a) : Required Area p p
10 p
ò ò = sin x dx - sin x dx
0 0
= 10 × 2 – 1 × 2 = 18 (Using period of |sin x| = p)
f(x) = |log|x|| p/4
ò 57. (b) : In = tan n x dx
0
1 (–1, 0) (1, 0) p/4
= 2ò log x dx òI n – 2 = tan n - 2 x dx
0 1 æ 0 p/4
0 çè p/4
( ) ò = ù1 1 ö
2 é x log x û0 - - x ÷ø . x dx ò ò \ I n + In – 2 = tan n x dx + tan n - 2 x dx
ë
= 2[(1 – 0) + (x)0 1] = 4 sq. units. 0 0
p/4 p/4
p x p x sin x ò ò = tan n - 2 x × (sec2 x –1) dx + tan n - 2 x dx
cos 2
ò ò 54. : 2 dx + 2 dx 0 0
(b) cos 2
-p 1 + x -p 1 + x
p/4
p x sin x p x sin x
+ cos 2 x + cos 2 x ò = tann - 2 x sec 2 x dx
ò ò = 0 + 2 2
1 dx = 2 × 0 1 0
-p
1
p x sin x p p sin x I n + In – 2 = n + 1
+ cos 2 x 2 0 1 + cos 2 x
ò ò = 4 dx dx 1
0 1 = 4 × \ n(In + In – 2) = 1 + 1/n
ò ò æ p xf x) pp ö \ Lt n(In + In – 2) = 1
ç by using (sin dx = f (sin x) dx÷
è0 2 ø n ® ¥
0 2 1 2
ò = 4 p p/2 ò ò ò 58. (c) : [x2 ]d x = [x2 ]dx + [x2 ] dx
2 ´ 2 × dx 0 0 1
sin x
0 1 + cos 2 x 2
= 4p (tan-1 cos x) 0p (By putting cos x = t) ò = 0 + 1 dx = 2 - 1
2 1
Differential Equations 63
CHAPTER DIFFERENTIAL EQUATIONS
11
1. Solution of the differential equation 7. If x dy = y(log y - log x + 1), then the solution of the equation
cosxdy = y(sinx – y)dx, 0 < x < p/2 is dx
(a) sec x = (tanx + c)y (b) y secx = tanx + c is
(c) y tanx = secx + c (d) tanx = (secx + c)y (2010) (a) x log æèç y ö = cy (b) y log æèç x ö = cx
x ø÷ y ÷ø
2. The differential equation which represents the family of curves
y = c e c 2x , where c and c are arbitrary constants, is (c) log èæç x ö = cy (d) log æ y ö = cx . (2005)
1 1 2 y ø÷ çè x ÷ø
(a) y¢¢ = y¢y (b) yy¢¢ = y¢
(c) yy¢¢ = (y¢) 2 (d) y¢ = y2 (2009)
8. The differential equation representing the family of curves
dy x + y y2 = 2c( x + c ), where c > 0, is a parameter, is of order and
3. The solution of the differential equation dx = x degree as follows
satisfying the condition y(1) = 1 is (a) order 1, degree 1 (b) order 1, degree 2
(c) order 2, degree 2 (d) order 1, degree 3. (2005)
(a) y = x ln x + x (b) y = ln x + x
(c) y = x ln x + x 2 (d) y = x e (x – 1) (2008) 9. The solution of the differential equation
4. The differential equation of the family of circles with fixed ydx + (x + x2 y )dy = 0 is
radius 5 units and centre on the line y = 2 is (a) 1 + log y = C (b) - 1 + log y = C
(a) (x – 2) 2 y¢2 = 25 – ( y – 2) 2 xy xy
(b) (x – 2) y¢2 = 25 – ( y – 2) 2 (c) - 1 = C (d) logy = Cx. (2004)
(c) ( y – 2) y¢ 2 = 25 – ( y – 2) 2 xy
(d) ( y – 2) 2 y¢ 2 = 25 – ( y – 2) 2 (2008) 10. The differential equation for the family of curves x2 + y2 – 2ay
5. The differential equation of all circles passing through the = 0, where a is an arbitrary constant is
origin and having their centres on the xaxis is (a) (x 2 – y 2) y¢ = 2xy (b) 2(x 2 + y 2) y¢ = xy
(c) 2(x 2 – y 2) y¢ = xy (d) (x 2 + y 2) y¢ = 2xy. (2004)
(a) y2 = x2 + 2 xy ddxy
(b) y2 = x2 - 2 xy ddxy 11. If x = e y + e y + ... to ¥ dy is
(c) x2 = y2 + xy ddyx dx
(d) x2 = y2 + 3xy ddyx . , x > 0 then
(a) 1 - x (b) 1
x x
(2007) (c) x (d) 1 + x . (2004)
1 + x x
6. The differential equation whose solution is Ax 2 + By2 = 1, 12. The solution of the differential equation
where A and B are arbitrary constants is of (1 + y2) + (x - e tan -1 y ) dy = 0 is
(a) second order and second degree dx
(b) first order and second degree
(c) first order and first degree (a) 2 xetan-1 y = e2 tan -1 y + k
(d) second order and first degree.
(b) xetan-1 y = tan -1 y + k
(2006) (c) xe2 tan-1 y = etan -1 y + k
(d) (x - 2) = ke- tan -1 y . (2003)
64 JEE MAIN CHAPTERWISE EXPLORER
13. The degree and order of the differential equation of the family (a) 1, 2 (b) 3, 1 (c) 3, 3 (d) 1, 2.
3 (2002)
of all parabolas whose axis is xaxis, are respectively
(a) 1, 2 (b) 3, 2 d 2 y
dx 2
(c) 2, 3 (d) 2, 1. (2003) 15. The solution of the equation = e -2 x
14. The order and degree of the differential equation (a) 1 e-2 x (b) 1 e-2 x + cx + d
4 4
( ) 1 dy 2 4 d 3 y
+ 3 dx 3 = d x 3 are
(c) 1 e-2 x + cx2 + d (d) 1 e-2 x + c + d . (2002)
4 4
Answer Key
1. (a) 2. (c) 3. (a) 4. (d) 5. (a) 6. (d)
7. (d) 8. (d) 9. (b) 10. (a) 11. (a) 12. (a)
13. (a) 14. (c) 15. (b)
Differential Equations 65
1. (a) : 1st solution: Þ v = ln x + ln k
cos x dy = y(sin x – y)dx As v = y/x we have y = x ln x + (ln k)x
At x = 1, y = 1 giving
Þ cos x dy = y sin x dx – y 2 dx 1 = 0 + (ln k) \ ln k = 1, Then y = x ln x + x
2nd Method (Inspection) :
Þ cos x dy – y sin x dx = –y 2 dx Rewriting the equation
Þ d(ycos x) = –y 2 dx Þ d(y cos x) = - codsx2 x
(y cos x)2
On integration, we have dy x + y
Þ – sec x = –y tan x + yk = as
dx x
Þ sec x = y(tanx + C) where C is a constant xdy – ydx = xdx
2nd solution: xdy - ydx dx
We have =
dy y(sin x - y) dy = y tan x - y2 sec x x2 x
dx = cos x Þ dx
d èæ y ö dx
dy - y tan x = -y2 sec x x ø = x
dx
Þ y
On integration x = ln x + k
1 dy - 1 tan x = - sec x
Þ y2 dx y Þ y = x ln x + kx
Setting, - 1 = v, we have As before, evaluating constant, y = x ln x + x
y 4. (d) : The equation to circle is
dv + (tan x)v = - sec x , which is linear in v. (x – a) 2 + (y – 2) 2 = 25 ...(1)
dx Differentiation w.r.t. x
I.F. = eò tan x dx = elnsec x = sec x ( x - a) + ( y - 2) dy = 0
dx
The solution is Þ x - a = -( y - 2) dy ...(2)
dx
v ´ sec x = ò - sec2 x dx + k
From (1) and (2) on eliminate ‘a’
Þ v sec x = - tan x + k
2)2 æ dy ö 2 2)2
Þ - sec x = - tan x - C Þ sec x = y(tan x + C) è dx ø
y ( y - + ( y - = 25
2. (c) : y = c1e c2 x Þ (y – 2) 2 (y¢)2 = 25 – (y – 2) 2
Differentiating w.r.t. x, we get 5. (a) : General equation of all such circles is
(x – h)2 + (y – 0)2 = h2 .... (i) where h is parameter
y¢ = c1c2ec2 x = c2 y ...(i) Þ (x – h)2 + y 2 = h 2
Again differentiating w.r.t. x ...(ii) Differentiating, we get 2( x - h) + 2 y ddyx = 0
y¢¢ = c2 y¢ h = x + y ddxy to eliminate h, putting value of h in
From (i) and (ii) upon division
y¢ = y Þ y¢¢y = ( y ¢)2 equation .... (i) ,
y¢¢ y ¢ \ we get y2 = x2 + 2xy ddxy .
Which is the desired differential equation of the family of
curves. 6. (d) : Given A x 2 + B y 2 = 1
3. (a) : 1st Method (Homogeneous equation): As solution having two constants, \ order of differential
dy + x ddvx equation is 2 so our choices (b) & (c) are discarded from the
dx
Let y = vx, so that = v list, only choices (a) and (b) are possible
dv x + vx Again A x2 + B y 2 = 1 .... (*)
We have v + x dx = x = 1 + v
Þ - BA = y dy .... (i)
x dx
dv dx Differentiating (*) w.r.t. x
Þ x dx = 1 Þ dv = x
Again on differentiating
66 JEE MAIN CHAPTERWISE EXPLORER
- BA = æ d 2 y ö æ dy ö2 ......(ii) 11. (a) : x = e y+e y+e y .... Þ x = e y + x
y ç dx2 ÷+ çè dx ÷ø
ø Differentiate w.r.t. x after taking logarithm both sides
è
By (i) and (ii) we get 1 1 + dy dy 1 - x
x = dx dx = x
d 2 y ö2 \ Þ
dx2 ÷
xy + x æ dy ø = y çèæ dy ö 12. (a) : From the given equation
çè dx dx ÷ø
Þ order 2 degree 1. (1 + y 2) dx + 1x = etan -1 y
dy
dy
7. (d) : x dx = y(log y - log x + 1) dx 1 e tan -1 y
ò Þ
+ 1 + y2 x = 1 + y2 Þ x ∙ I.F. = y × I.F. dy
dy
dy y æ æ y ö ö y
\ dx = x çè log çè x ø÷ + 1 ÷ø Now put x = v 1
1+ y 2
òe dy etan -1 y
\ v logv dx = x dv where I.F = =
Þ dv = dx Þ log æ y ö = cx . Þ x etan -1 y = e 2 tan -1 y + c
v log v x çè x ÷ø 2
8. (d) : y2 = 2 c( x + c ) ...(i) Þ x etan-1 y = e2 tan -1 y + k y
\ 2yy 1 = 2c \ yy1 = c 13. (a): As axis of parabola is xaxis which
Now putting c = yy1 in (i) we get
means focus lies on xaxis. Equation of such x
( )y2 = 2 × yy1 x + yy1 Þ ( y2 - 2 xyy1 )2 = 4 ( yy1 ) 3 parabola’s is given by (a , 0)
y 2 = 4a(x – k) ... (i)
Þ (y 2 – 2xyy 1) 2 = 4y 3y 1 3 Þ order 1, degree 3. Þ 2yy1 = 4a (by differentiating (i) w.r.t. x)
9. (b) : y dx = –(x 2y + x) dy Þ y dy = 2a ... (ii)
dx
Þ ydx + xdy = –x2 y dy
d 2 y dy ö 2
ydx + xdy -dy d (xy) dy Þ dx2 + æ dx ÷ø = 0
(xy) 2 = y (xy) 2 = - y èç
Þ Þ
(by differentiating (ii) w.r. to x)
Þ d çæè - 1 ö = - dy Þ order 2 and degree 1 (Concept: Exponent of highest order
xy ÷ø y derivative is called degree and order of that derivative is called
Þ - 1 = –log y + C order of the differential equation.)
xy Þ 2 x etan - 1 y = e2 tan -1 y + k
2 æ d 3 y ö
4 ççè dx3 ÷÷ø
1 14. (c) : æ 1 + 3 dy ö 3 =
Þ - + log y = C çè dx ÷ø
xy é d 3 y ù3
10. (a) : Given family of curve is Þ çæè1 + 3 dy ö 2 = ê 4 dx3 ú
x 2 + y2 – 2ay = 0 dx ø÷ ë û
...(1)
x2 + y 2 \ highest order is 3 whose exponent is also 3.
Þ 2a = y
d 2 y
Also from (1), 2x + 2yy¢ – 2a y¢ = 0 15. (b) : Given dx 2 = e– 2x
æ x2 + y 2 ö dy e-2 x
Þ 2x + 2yy¢ – ççè y ÷÷ø y¢ = 0 \ = + c
Þ 2xy + y¢(2y 2 – x2 – y 2 ) = 0 Þ y¢(x2 – y2 ) = 2xy
dx -2
\ y = e- 2 x + cx + d
4
Two Dimensional Geometry 67
CHAPTER TWO DIMENSIONAL
GEOMETRY
12
1. The circle passi ough (1, –2) and touching the axis of x at 7. Statement 1 : An equation of a common tangent to the parabola
(3, 0) also passes through the point y2 = 16 3 x and the ellipse 2x2 + y2 = 4 is y = 2x + 2 3
Statement 2 : If the line y = mx + 4 3 , (m ¹ 0) is a common
(a) (2, –5) (b) (5, –2)
m
(c) (–2, 5) (d) (–5, 2) (2013)
2. Given : A circle, 2x 2 + 2y 2 = 5 and a parabola y2 = 4 5x . tangent to the parabola y2 = 16 3 x the ellipse 2x 2 + y 2 = 4,
Statement1 : An equation of a common tangent to these then m satisfies m4 + 2m 2 = 24.
curves is y = x + 5. (a) Statement 1 is true, Statement 2 is true; Statement 2 is
Statement2 : If the line, y = mx + 5 (m ¹ 0) is their not a correct explanation for Statement 1.
m (b) Statement 1 is true, Statement 2 is false.
(c) Statement 1 is false, Statement 2 is true.
common tangent, then m satisfies m 4 – 3m 2 + 2 = 0.
(a) Statement1 is true, Statement2 is true, Statement2 is (d) Statement 1 is true, Statement 2 is true; Statement 2 is
not a correct explanation for Statement1. a correct explanation for Statement 1. (2012)
(b) Statement1 is true, Statement2 is false. 8. The length of the diameter of the circle which touches the
(c) Statement1 is false, Statement2 is true. xaxis at the point (1, 0) and passes through the point (2, 3) is
(d) Statement1 is true, Statement2 is true, Statement2 is a (a) 6/5 (b) 5/3
correct explanation for Statement1. (2013) (c) 10/3 (d) 3/5 (2012)
3. A ray of light along x + 3y = 3 gets reflected upon reaching 9. An ellipse is drawn by taking a diameter of the circle
x-axis, the equation of the reflected ray is (x – 1) 2 + y2 = 1 as its semiminor axis and a diameter of the
(a) 3y = x - 3 (b) y = 3x - 3 circle x 2 + (y – 2) 2 = 4 as its semimajor axis. If the centre
(c) 3y = x - 1 (d) y = x + 3 (2013) of the ellipse is at the origin and its axes are the coordinate
4. The equation of the circle passing through the focii of the axes, then the equation of the ellipse is
(a) 4x 2 + y 2 = 8 (b) x2 + 4y2 = 16
ellipse x2 + y2 = 1 and having centre at (0, 3) is (c) 4x 2 + y 2 = 4 (d) x2 + 4y 2 = 8 (2012)
16 9
10. A line is drawn through the point (1, 2) to meet the coordinate
(a) x2 + y 2 – 6y + 7 = 0 (b) x2 + y 2 – 6y – 5 = 0
axes at P and Q such that it forms a triangle OPQ, where O
(c) x2 + y2 – 6y + 5 = 0 (d) x 2 + y2 – 6y – 7 = 0 is the origin. If the area of the triangle OPQ is least, then the
(2013) slope of the line PQ is
5. The xcoordinate of the incentre of the triangle that has the (a) – 2 (b) – 1/2
coordinates of mid points of its sides as (0, 1), (1, 1) and (1, 0) (c) – 1/4 (d) – 4 (2012)
is (b) 1 + 2 11. The two circles x 2 + y2 = ax and x2 + y 2 = c2 ( c > 0) touch each
(a) 2 - 2 other if
(c) 1 - 2 (d) 2 + 2 (2013) (a) a = 2c (b) |a| = 2c
6. If the line 2x + y = k passes through the point which divides (c) 2 |a| = c (d) |a| = c (2011)
the line segment joining the points (1, 1) and (2, 4) in the 12. The lines L 1 : y – x = 0 and L2 : 2x + y = 0 intersect the line
ratio 3 : 2, then k equals L3 : y + 2 = 0 at P and Q respectively. The bisector of the
(a) 6 (b) 11/5 (c) 29/5 (d) 5 acute angle between L 1 and L 2 intersects L3 at R.
(2012) Statement1 : The ratio PR : RQ equals
68 JEE MAIN CHAPTERWISE EXPLORER
Statement2 : In any triangle, bisector of an angle divides 21. Three distinct points A, B and C are given in the 2dimensional
coordinate plane such that the ratio of the distance of any
the triangle into two similar triangles. one of them from the point (1, 0) to the distance from the
(a) Statement1 is true, Statement2 is false. point ( –1, 0) is equal to 1/3. Then the circumcentre of the
(b) Statement1 is false, Statement2 is true. triangle ABC is at the point
(c) Statement1 is true, Statement2 is true; Statement2 is a
( ) (a) 45 , 0 ( ) (b) 52 , 0
correct explanation for Statement1.
(d) Statement1 is true, Statement2 is true; Statement2 is
not a correct explanation for Statement1. (2011)
13. The shortest distance between line y – x = 1 and curve ( ) 5
x = y 2 is (c) 3 , 0 (d) (0, 0) (2009)
8 (b) 4 (c) 3 (d) 32 22. The ellipse x 2 + 4y 2 = 4 is inscribed in a rectangle aligned
(a) 3 4 8
with the coordinate axes, which in turn is inscribed in another
3 2
(2011)
14. Equation of the ellipse whose axes are the axes of coordinates ellipse that passes through the point (4, 0). Then the equation
and which passes through the point (–3, 1) and has
of the ellipse is
(a) x2 + 12y 2 = 16 (b) 4x 2 + 48y 2 = 48
eccentricity 2 is (c) 4x 2 + 64y 2 = 48 (d) x2 + 16y 2 = 16 (2009)
5
23. If P and Q are the points of intersection of the circles
(a) 3x 2 + 5y 2 – 15 = 0 (b) 5x 2 + 3y 2 – 32 = 0
(d) 5x 2 + 3y 2 – 48 = 0 (2011) x2 + y 2 + 3x + 7y + 2p – 5 = 0 and x 2 + y 2 + 2x + 2y – p 2 = 0,
(c) 3x 2 + 5y 2 – 32 = 0
4 then there is a circle passing through P, Q and (1, 1) for
x2
15. The equation of the tangent to the curve y = x + , that is (a) all except one value of p
parallel to the xaxis, is (b) all except two values of p
(a) y = 0 (b) y = 1 (c) exactly one value of p
(c) y = 2 (d) y = 3 (2010) (d) all values of p (2009)
16. If two tangents drawn from a point P to the parabola 24. A focus of an ellipse is at the origin. The directrix is the line
y2 = 4x are at right angles, then the locus of P is
1
(a) x = 1 (b) 2x + 1 = 0 x = 4 and the eccentricity is . Then the length of the semi
2
(c) x = –1 (d) 2x – 1 = 0 (2010) major axis is
17. The line L given by x + y = 1 passes through the point 5 8 2 4
5b (a) 3 (b) 3 (c) 3 (d) 3
(2008)
(13, 32). The line K is parallel to L and has the equation 25. The point diametrically opposite to the point P(1, 0) on the
circle x 2 + y 2 + 2x + 4y – 3 = 0 is
x + y = 1 . Then the distance between L and K is
c 3
(a) (3, 4) (b) (3, – 4)
23 17 23
(a) 15 (b) 17 (c) 15 (d) 17 (c) (– 3, 4) (d) (– 3, – 4) (2008)
(2010) 26. A parabola has the origin as its focus and the line
x = 2 as the directrix. Then the vertex of the parabola is at
18. The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x – 4y = m (a) (2, 0) (b) (0, 2) (c) (1, 0) (d) (0, 1)
(2008)
at two distinct points if
(a) –85 < m < –35 (b) –35 < m < 15
(c) 15 < m < 65 (d) 35 < m < 85 (2010) 27. The perpendicular bisector of the line segment joining
19. The shortest distance between the line y – x = 1 and the curve P(1, 4) and Q(k, 3) has yintercept – 4. Then a possible value
x = y 2 is
of k is
(a) 2 3 (b) 3 2 (c) 3 (d) 3 2 (a) – 4 (b) 1 (c) 2 (d) – 2
8 5 4 8
(2009) (2008)
28. The normal to a curve at P(x, y) meets the xaxis at G. If the
20. The differential equation which represents the family of curves distance of G from the origin is twice the abscissa of P, then
y = c1e c2 x , where c1 and c2 are arbitrary constants, is the curve is a
(a) y¢¢ = y¢y (b) yy¢¢ = y¢ (a) circle (b) hyperbola
(c) yy¢¢ = (y¢) 2 (d) y¢ = y 2 (2009) (c) ellipse (d) parabola. (2007)
Two Dimensional Geometry 69
29. Consider a family of circles which are passing through the 37. If the lines 3x – 4y – 7 = 0 and 2x – 3y – 5 = 0 are two diameters
point (–1, 1) and are tangent to xaxis. If (h, k) are the coordinate of a circle of area 49p square units, then the equation of the
of the centre of the circles, then the set of values of k is circle is
given by the interval (a) x2 + y 2 + 2x – 2y – 47 = 0
(a) - 1 £ k £ 1 (b) k £ 12 (b) x2 + y 2 + 2x – 2y – 62 = 0
2 2
(c) x 2 + y 2 – 2x + 2y – 62 = 0
(c) 0 £ k £ 1 (d) k ³ 21 . (2007) (d) x 2 + y 2 – 2x + 2y – 47 = 0. (2006)
2
38. In an ellipse, the distance between its focii is 6 and minor axis
30. If one of the lines of my 2 + (1 – m 2 )xy – mx 2 = 0 is a bisector is 8. Then its eccentricity is,
of the angle between the lines xy = 0, then m is (a) 3/5 (b) 1/2
(a) 1 (b) 2 (c) –1/2 (d) –2. (c) 4/5 (d) 1/ 5 . (2006)
(2007) 39. The locus of the vertices of the family of parabolas
31. Let P = (–1, 0), Q = (0, 0) and R = (3, 3 3) be three points. y = a3x2 + a2 x - 2 a is (b) xy = 3
The equation of the bisector of the angle PQR is 3 2 4
(a) 3 x + y = 0 (b) x + 3 y = 0 (a) xy = 105
2 64
(c) 3x + y = 0 (d) x+ 3 y = 0. (2007) (c) xy = 35 (d) xy = 64 . (2006)
2 16 105
32. Let A(h, k), B(1, 1) and C(2, 1) be the vertices of a right 40. A straight line through the point A(3, 4) is such that its intercept
angled triangle with AC as its hypotenuse. If the area of the between the axes is bisected at A. Its equation is
triangle is 1 square unit, then the set of values which ‘k’ can (a) x + y = 7 (b) 3x – 4y + 7 = 0
take is given by (c) 4x + 3y = 24 (d) 3x + 4y = 25. (2006)
(a) {–1, 3} (b) {–3, –2} 41. If the pair of lines ax2 + 2(a + b)xy + by2 = 0 lie along
diameters of a circle and divide the circle into four sectors
(c) {1, 3} (d) {0, 2}. (2007) such that the area of one of the sectors is thrice the area of
another sector then
33. The equation of a tangent to the parabola y 2 = 8x is (a) 3a 2 – 2ab + 3b 2 = 0 (b) 3a2 – 10ab + 3b 2 = 0
(c) 3a 2 + 2ab + 3b 2 = 0 (d) 3a2 + 10ab + 3b 2 = 0.
y = x + 2. The point on this line from which the other tangent (2005)
to the parabola is perpendicular to the given tangent is
(a) (2, 4) (b) (–2, 0)
(c) (–1, 1) (d) (0, 2). (2007) 42. The locus of a point P (a, b) moving under the condition that the
34. For the hyperbola x2 - y 2 = 1, which of the following line y = ax + b is a tangent to the hyperbola x2 - y 2 = 1 is
cos2 sin 2 a a2 b2
a (a) a circle (b) an ellipse
remains constant when a varies ?
(a) abscissae of vertices (b) abscissae of foci (c) a hyperbola (d) a parabola. (2005)
(c) eccentricity (d) directrix. (2007) 43. An ellipse has OB as semi minor axis, F and F ¢ its focii and
the angle FBF ¢ is a right angle. Then the eccentricity of the
35. If (a, a 2) falls inside the angle made by the lines y = x , x > 0 ellipse is
2
(a) 1 1 1 (d) 14 . (2005)
and y = 3x, x > 0, then a belongs to 2 (b) 2 (c) 3
æ 1 ö 44. If a circle passes through the point (a, b) and cuts the circle
çè 2 ÷ø
(a) 0, (b) (3, ¥) x2 + y2 = p 2 orthogonally, then the equation of the locus of
æ 1 , 3 ÷øö æ 1 ö its centre is
çè 2 çè 2 ÷ø .
(c) (d) -3, - (2006) (a) 2ax + 2by – (a2 – b2 + p 2) = 0
36. Let C be the circle with centre (0, 0) and radius 3 units. The (b) x2 + y 2 – 3ax – 4by + (a 2 + b2 – p 2) = 0
equation of the locus of the mid points of chord of the circle (c) 2ax + 2by – (a2 + b 2 + p 2) = 0
C that subtend an angle of 2p/3 at its centre is (d) x2 + y 2 – 2ax – 3by + (a 2 – b 2 – p 2) = 0. (2005)
(a) x2 + y2 = 3 45. A circle touches the xaxis and also touches the circle with centre
2
(b) x2 + y 2 = 1 at (0, 3) and radius 2. The locus of the centre of the circle is
27 9 (a) a circle (b) an ellipse
= = .
(c) x2 + y2 4 (d) x2 + y2 4 (2006) (c) a parabola (d) a hyperbola. (2005)
70 JEE MAIN CHAPTERWISE EXPLORER
46. If the circles x 2 + y 2 + 2ax + cy + a = 0 and 54. If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along
diameters of a circle of circumference 10p, then the equation
x2 + y 2 – 3ax + dy – 1 = 0 intersect in two distinct points P of the circle is
and Q then the line 5x + by – a = 0 passes through P and Q for
(a) no value of a (a) x2 + y2 + 2x + 2y – 23 = 0
(b) exactly one value of a (b) x 2 + y 2 – 2x – 2y – 23 = 0
(c) exactly two values of a (c) x 2 + y 2 – 2x + 2y – 23 = 0
(d) infinitely many values of a. (2005)
(d) x 2 + y 2 + 2x – 2y – 23 = 0. (2004)
47. If a vertex of a triangle is (1, 1) and the mid points of two 55. A variable circle passes through the fixed point
sides through this vertex are (–1, 2) and (3, 2), then the centroid A(p, q) and touches xaxis. The locus of the other end of the
of the triangle is diameter through A is
( ) (a)-3 1, 7 ( ) (b) 7 (a) (y – p)2 = 4qx (b) (x – q)2 = 4py
3 -1 , 3
( ) (c) ( ) (d) (c) (x – p)2 = 4qy (d) (y – q)2 = 4px. (2004)
1 , 7 1, 7 . (2005) 56. If a circle passes through the point (a, b) and cuts the circle
3 3 3 x2 + y2 = 4 orthogonally, then the locus of its centre is
48. If nonzero numbers a, b, c are in H.P., then the straight line (a) 2ax – 2by + (a2 + b2 + 4) = 0
x + y + 1 = 0 always passes through a fixed point. That point (b) 2ax + 2by – (a2 + b 2 + 4) = 0
a b c
is (c) 2ax + 2by + (a 2 + b2 + 4) = 0
(a) (–1, –2) (b) (–1, 2) (d) 2ax – 2by – (a 2 + b 2 + 4) = 0. (2004)
( ) (c)
1, - 1 (d) (1, –2). (2005) 57. If one of the lines given by 6x 2 – xy + 4cy2 = 0 is 3x + 4y = 0,
2
49. The line parallel to the xaxis and passing through the intersection then c equals
of the lines ax + 2by + 3b = 0 and bx – 2ay – 3a = 0, where
(a, b) ¹ (0, 0) is (a) 3 (b) – 1 (c) 1 (d) – 3.
(a) below the xaxis at a distance of 2/3 from it
(b) below the xaxis at a distance of 3/2 from it (2004)
(c) above the xaxis at a distance of 2/3 from it
(d) above the xaxis at a distance of 3/2 from it. (2005) 58. If the sum of the slopes of the lines given by x2 – 2cxy – 7y2 = 0
is four times their product, then c has the value
(a) 2 (b) –1 (c) 1 (d) –2.
(2004)
50. Let P be the point (1, 0) and Q a point on the locus y2 = 8x. 59. The equation of the straight line passing through the point
(4, 3) and making intercepts on the coordinate axes whose
The locus of mid point of PQ is sum is –1 is
(a) x2 – 4y + 2 = 0 (b) x 2 + 4y + 2 = 0
(c) y 2 + 4x + 2 = 0 (d) y 2 – 4x + 2 = 0. (2005) (a) x + y = 1 and x + y = 1
2 3 2 1
51. The eccentricity of an ellipse, with its centre at the origin, is
1/2. If one of the directrices is x = 4, then the equation of the (b) x - y = - 1 and x + y = 1
2 3 - 2 1
ellipse is
(a) 4x 2 + 3y2 = 12 (b) 3x 2 + 4y2 = 12 (c) x + y = - 1 and x + y = - 1
2 3 - 2 1
(c) 3x2 + 4y 2 = 1 (d) 4x 2 + 3y2 = 1. (2004)
x y x y
52. If a ¹ 0 and the line 2bx + 3cy + 4d = 0 passes through the (d) 2 - 3 = 1 and - 2 + 1 = 1. (2004)
points of intersection of the parabolas y2 = 4ax and x 2 = 4ay,
then 60. Let A(2, –3) and B(–2, 1) be vertices of a triangle ABC. If the
(a) d 2 + (2b – 3c)2 = 0 (b) d 2 + (3b + 2c) 2 = 0
(c) d 2 + (2b + 3c)2 = 0 (d) d 2 + (3b – 2c)2 = 0. centroid of this triangle moves on the line 2x + 3y = 1, then
(2004)
the locus of the vertex C is the line
(a) 3x + 2y = 5 (b) 2x – 3y = 7
53. The intercept on the line y = x by the circle x2 + y 2 – 2x = 0 (c) 2x + 3y = 9 (d) 3x – 2y = 3. (2004)
is AB. Equation of the circle on AB as a diameter is
(a) x2 + y2 + x + y = 0 (b) x2 + y2 – x + y = 0 61. The normal to the curve x = a(1 + cosq), y = a sinq at q always
(c) x 2 + y2 – x – y = 0 (d) x2 + y 2 + x – y = 0.
(2004) passes through the fixed point
(a) (0, 0) (b) (0, a)
(c) (a, 0) (d) (a, a). (2004)
Two Dimensional Geometry 71
62. A point on the parabola y 2 = 18x at which the ordinate increases 69. The normal at the point (bt 12 , 2bt 1 ) on a parabola meets the
parabola again in the point (bt 22 , 2bt2 ), then
at twice the rate of the abscissa is
( ) (a)- 9, 9 (b) (2, –4) (a) t2 = -t1 + t21 (b) t2 = t1 - t21
8 2
(c) (2, 4) ( ) (d) 9 , 9 (2004) (c) t2 = t1 + t21 (d) t2 = -t1 - t21 .
8 2
(2003)
63. If the equation of the locus of point equidistant from the points
(a1 , b 1) and (a 2, b2 ) is 70. The foci of the ellipse 1x6 2 + y 2 = 1 and the hyperbola
(a1 – a2 )x + (b 1 – b2 )y + c = 0, then c = b2
(a) a12 - a22 + b12 - b22 x2 y 2
144 - 81 = 215 coincide. Then the value of b 2 is
(b) 12 (a12 + a22 + b12 + b22 ) (a) 5 (b) 7 (c) 9 (d) 1.
(2003)
(c) (a12 + b12 - a22 - b22 ) 71. A triangle with vertices (4, 0), (–1, –1), (3, 5) is
(d) 12 (a22 + b22 - a12 - b12 ). (a) isosceles and right angled
(b) isosceles but not right angled
(2003) (c) right angled but not isosceles
(d) neither right angled nor isosceles
64. Locus of centroid of the triangle whose vertices are (2002)
(a cost, a sint), (b sint, –b cost) and (1, 0), where t is a parameter, 72. The equation of a circle with origin as a centre and passing
is through equilateral triangle whose median is of length 3a is
(a) (3x – 1) 2 + (3y)2 = a 2 + b 2 (a) x2 + y2 = 9a 2 (b) x2 + y2 = 16a 2
(b) (3x + 1) 2 + (3y) 2 = a 2 + b 2 (c) x2 + y2 = 4a2 (d) x 2 + y2 = a2 . (2002)
(c) (3x + 1) 2 + (3y) 2 = a 2 – b 2 73. The centre of the circle passing through (0, 0) and (1, 0) and
(d) (3x – 1) 2 + (3y) 2 = a 2 – b 2. (2003) touching the circle x2 + y2 = 9 is
65. If the pairs of straight lines x 2 – 2pxy – y 2 = 0 and ( ) (a) 1 , 1 ( ) (b) 12 , - 2
2 2
( ) (c) ( ) (d)
x 2 – 2qxy – y 2 = 0 be such that each pair bisects the angle 3 , 1 1 , 3 (2002)
2 2 2 2
between the other pair, then
(a) p = –q (b) pq = 1 74. Locus of mid point of the portion between the axes of
(c) pq = –1 (d) p = q. (2003) x cosa + y sina = p where p is constant is
66. A square of side a lies above the x axis and has one vertex (a) x2 + y 2 = 4 (b) x2 + y 2 = 4p 2
p 2
at the origin. The side passing through the origin makes an
1 1 2 1 1 4
angle a (0 < a < p/4) with the positive direction of x axis. (c) x2 + y2 = p 2 (d) x2 + y2 = p 2 . (2002)
The equation of its diagonal not passing through the origin is
(a) y (cosa + sina) + x (sina – cosa) = a 75. The point of lines represented by
(b) y (cosa + sina) + x (sina + cosa) = a 3ax2 + 5xy + (a2 – 2)y2 = 0
(c) y (cosa + sina) + x (cosa – sina) = a and ^ to each other for
(d) y (cosa – sina) – x (sina – cosa) = a. (2003) (a) two values of a (b) " a
(c) for one value of a (d) for no values of a.
67. The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle (2002)
having area as 154 sq. units. Then the equation of the circle is 76. The centres of a set of circles, each of radius 3, lie on the circle
(a) x 2 + y 2 + 2x – 2y = 47 x2 + y2 = 25. The locus of any point in the set is
(b) x 2 + y 2 – 2x + 2y = 47 (a) 4 £ x 2 + y 2 £ 64 (b) x 2 + y 2 £ 25
(c) x 2 + y 2 – 2x + 2y = 62 (c) x 2 + y 2 ³ 25 (d) 3 £ x2 + y 2 £ 9 (2002)
(d) x 2 + y 2 + 2x – 2y = 62. (2003) 77. If the pair of lines
68. If the two circles (x – 1) 2 + (y – 3) 2 = r 2 and ax2 + 2hxy + by 2 + 2gx + 2fy + c = 0
x2 + y2 – 8x + 2y + 8 = 0 intersect in two distinct points, then
intersect on the yaxis then
(a) r < 2 (b) r = 2 (a) 2fgh = bg2 + ch 2 (b) bg2 ¹ ch2
(c) r > 2 (d) 2 < r < 8.
(2003) (c) abc = 2fgh (d) none of these (2002)
72 JEE MAIN CHAPTERWISE EXPLORER
78. If the chord y = mx + 1 of the circle x2 + y 2 = 1 subtends an 79. Two common tangents to the circle x2 + y 2 = 2a 2 and parabola
y 2 = 8ax are
angle of measure 45° at the major segment of the circle then
value of m is (a) x = ±(y + 2a) (b) y = ±(x + 2a)
(a) 2 ± 2 (b) -2 ± 2 (c) x = ±(y + a) (d) y = ±(x + a). (2002)
(c) -1 ± 2 (d) none of these (2002)
Answer Key
1. (b) 2. (a) 3. (a) 4. (d) 5. (a) 6. (a)
7. (d) 8. (c) 9. (b) 10. (a) 11. (d) 12. (a)
13. (c) 14. (c) 15. (d) 16. (c) 17. (d) 18. (b)
19. (d) 20. (c) 21. (a) 22. (a) 23. (a) 24. (b)
25. (d) 26. (c) 27. (a) 28. (b, c) 29. (d) 30. (a)
31. (c) 32. (a) 33. (b) 34. (b) 35. (c) 36. (d)
37. (d) 38. (a) 39. (a) 40. (c) 41. (c) 42. (c)
43. (b) 44. (c) 45. (c) 46. (a) 47. (d) 48. (d)
49. (b) 50. (d) 51. (b) 52. (c) 53. (c) 54. (c)
55. (c) 56. (b) 57. (d) 58. (a) 59. (d) 60. (c)
61. (c) 62. (d) 63. (d) 64. (a) 65. (c) 66. (c)
67. (b) 68. (d) 69. (d) 70. (b) 71. (a) 72. (c)
73. (b) 74. (d) 75. (a) 76. (a) 77. (a) 78. (c)
79. (b)
Two Dimensional Geometry 73
1. (b) : The system of circles touches the line y = 0 at the point 6. (a) : A (1, 1) ; B (2, 4)
(3, 0) is given by {(x – 3) 2 + y 2 )} + ly = 0
As the circle passes through (1, –2), we can determine l which P(x1 , y1 ) divides line segment AB in the ratio 3 : 2
gives 4 + 4 – 2l = 0 \ l = 4 x1 = 3(2) + 2(1) = 8 y1 = 3( 4) + 2(1) = 14
The circle is (x – 3) 2 + y 2 + 4y = 0. A simple calculation 5 5 5 5
shows that (5, –2) lies on the circle. 2x + y = k passes through P(x 1 , y 1 )
2. (a) : Let a tangent to the parabola be \ 2 ´ 8 + 14 = k Þ k = 6
55
y = mx + 5 (m ¹ 0)
m 7. (d) : Statement 1 :
As it is a tangent to the circle x 2 + y 2 = 5/2, we have
æ 5 ö = 5 1 + m2 Þ (1 + m2 )m2 = 2 y2 = 16 3x , y = m x + 4 3
çè m ÷ø 2 m
which gives m 4 + m 2 – 2 = 0 Þ (m2 – 1)(m 2 + 2) = 0 x2 + y2 = 1 , x = m1y + 4 m12 + 2
2 4
As m Î R, m 2 = 1 \ m = ± 1
Also m = ± 1 does satisfy m 4 – 3m 2 + 2 = 0 Þ y= x - 4 + m21 2 m = 1
m1 m1
Hence common tangents are
y = x + 5 and y = - x - 5
3. (a) : As the slope of incident ray is - 1 so the slope of reflected Now, æ 43 ö 2 = æ 4 + 2 ö 2
3 çè m ÷ø çèç - m1 2 ÷ø÷
ray has to be 1 . Þ 48 = 4+ 2 = 4 + 2m2 Þ 24 = 2 + m2
3 m2 m12 m2
The point of incidence is ( 3 , 0) . Hence the equation of Þ m 4 + 2m 2 – 24 = 0 ...(1)
reflected ray is y = 1 (x - 3) . Þ (m 2 + 6) (m 2 – 4) = 0 Þ m = ± 2
3
\ 3y - x = - 3. \ x - 3y - 3 = 0
4. (d) : Foci are given by (± ae, 0) Statement 2 : If y = mx + 4 3 is a common tangent t o
As a2 e 2 = a2 – b2 = 7 we have equation of circle as m
(x – 0) 2 + (y – 3) 2 = ( 7 - 0)2 + (0 - 3)2 y2 = 16 3 x and ellipse 2x 2 + y 2 = 4, then m satisfies
\ x 2 + y 2 – 6y – 7 = 0
m 4 + 2m2 – 24 = 0
From (1), statement 2 is a correct explanation for statement 1.
5. (a) : The triangle whose 8. (c) : Let the equation of the circle is (x – 1) 2 + (y – k) 2 = k 2
It passes through (2, 3)
sides midpoints
\ 1 + 9 + k 2 – 6k = k 2
are given to be
Þ k = 5 Þ diameter = 10
(0, 1), (1, 0) and (1, 1) 33
happen to be a right 9. (b) :(x – 1) 2 + y2 = 1, r = 1 Þ a = 2
and x 2 + (y – 2) 2 = 4, r = 2 Þ b = 4
angled triangle with Þ x2 + y2 = 1 Þ x2 + 4y2 = 16
16 4
vertices as shown.
1 st solution : xcoordinate of incentre
= ax1 + bx2 + cx3 = 2 ´ 2 + 2 2 ´ 0 + 2 ´ 0
a+b+c 2 + 2 + 2 2
= 4 = 2 = 2 - 2 10. (a) : y = mx + c Þ 2 = m + c
4 + 2 2 2 + 2 Coordinates of P & Q : P(0, c), Q(–c/m, 0)
2n d solution : r = (s - a)tan A 1 ´|c| ´ c c2 = A
2 2 m 2m
æ4 ö = A Þ
èç +2 2 - 2 ÷ø p
= 2 2 ta n 4 = 2- 2
74 JEE MAIN CHAPTERWISE EXPLORER
Þ (2 - m) 2 = A Þ m2 - 4m + 4 = A Þ m - 2 + 2 = A 16. (c) : From a property of the parabola, the perpendicular tangents
2m 2m 2 m
intersect at the directrix.
Q dA = 0 The equation of directrix is x = –1, hence this is the locus of point
dm
P.
1 2 1 2 Þ m 2 = 17. (d) : As the line passes through (13, 32), we have
2 m2 2 m2
Þ - = 0 Þ = 4 Þ m = ± 2 13 + 32 = 1 Þ 32 = 1 - 13 = - 8 Þ b = - 20
11. (d) : The centres and radii are 5b b 55
èçæ x a ø÷ö 2 + y2 a2 , x2 + y2 = c2 Thus the line is x - y = 1, i.e., 4x - y = 20
2 4 5 20
- =
The equation of line parallel to 4x – y = 20 has slope 4.
Centre æèç a , 0÷öø and (0, 0) & radius = a and c Thus - 3 = 4. \ c = - 3 .
2 2 c 4
æ a ø÷ö 2 + (0 - 0) = a ±c Þ a = a ±c Then the equation to line k is 4x – y = –3
èç 2 2 22
The distance between lines k and c is
Þ a = c- a . \ |a| = c. 20 + 3 = 23
2 2 42 + 12 17
18. (b) : The circle is x2 + y 2 – 4x – 8y – 5 = 0
12. (a) : In triangle OPQ, O divides PQ in the ratio of OP : OQ
which is 2 2 : 5 but it fails to divide triangle into two similar Þ (x – 2) 2 + (y – 4) 2 = 5 2
triangles. Length of perpendicular from centre (2, 4) on the line 3x – 4y
13. (c) : Let P be (y 2, y) – m = 0 should be less than radius.
Perpendicular distance from P to x – y + 1 = 0 is Þ |6 - 16 - m| < 5 Þ (10 + m) < 25
|y2 - y + 1| 5
2 Þ –25 < 10 + m < 25 Þ –35 < m < 15
As |y 2 – y + 1| = y 2 – y + 1 (Q y2 – y + 1 > 0 )
1 × (4 ac - b2 ) 19. (d) :
2 4 a
Minimum value =
= 1 × 4 - 1 = 3
2 4 × 1 4 2
14. (c) : Let the ellipse be x 2 + y 2 = 1
a2 b2
(–3, 1) lies on it Þ 9 + 1 = 1
a2 b2
Let P(t 2, t) be a point on y 2 = x
Also b2 = a2 çèæ1 - 52 ø÷ö Þ 5b2 = 3a 2 The slope of normal at P
a2 32 b2 32 = - the slope 1 at P = - 1 1 = - 2t
7 5 of tangent / 2t
Upon solving we get = , =
The equation to ellipse becomes The shortest distance between the curves occur along the common
3x 2 + 5y 2 = 32 normal. Thus –2t = –1
Þ t = 1/2
15. (d) : y = x + 4 Thus the point P is (1/4, 1/2)
x2
The required shortest distance
On differentiation, dy = 1 - 8 ( ) =1 - 1 + 1 3 3 2
dx x3 4 2 42 8
= =
As the tangent is parallel to xaxis, we have 2
1- 8 = 0 Þ x3 = 8. \ x = 2 20. (c) : y = c1e c2 x
x3
Differentiating w.r.t. x, we get
So, y = 2 + 4 = 2 + 1 = 3 ...(i)
22 y¢ = c1c2ec2 x = c2 y
Thus (2, 3) is the point of contact and equation of the tangent Again differentiating w.r.t. x
is y = 3. y¢¢ = c2y¢ ....(ii)
Two Dimensional Geometry 75
From (i) and (ii) upon division Þ a - ae = 4 (note that ae > ae )
e
y¢ y
y¢¢ = y¢ Þ y¢¢y = ( y¢ ) 2 Þ a æè 1 e øö = 4 Þ a æ 2 - 1 ö = 4
wh ich is the desired differen tial equ ation of the family of cu rves. e- è 2 ø
21. (a) : Let P be a general point (x, y) such that Þ a × 3 = 4 \ a = 8
2 3
PM 1
PN = 3 where M º (1, 0) and N º (–1, 0) Remark : The question should have read “The corresponding
directrix” in place of “the directrix”.
(x - 1) 2 + y 2 1
we have (x + 1) 2 + y2 = 3 25. (d) : The centre C of the circle is seen to be
Þ 9[(x – 1) 2 + y 2] = (x + 1) 2 + y 2
which reduces to (– 1, – 2). As C is the mid point P and P¢, the coordinate of
8x 2 + 8y 2 – 20x + 8 = 0
P¢ is given by
Þ x2 + y 2 - 10 + 1 = 0 P¢ º (2 × – 1 – 1, 2 × – 2 – 0) P C P¢
4 x º (– 3, – 4) (1, 0) (–1, –2)
Remark : If P be (a, b) and C(h, k)
Þ x2 + y 2 - 5 +1 = 0 then
2 x
P¢ º (2h – a, 2k – b)
The locus is a circle with centre (5/4, 0) 26. (c) : The vertex is F V (2, 0)
As points A, B, C lie on this circle, the circumcentre of triangle the mid point of FN, (0, 0) N
ABC is (5/4, 0). that is, vertex = (1, 0)
22. (a) : The given ellipse is x2 + y 2 = 1 27. (a) : The slope of x = 2
4 1
l = - the slope of the 1o riginal line PQ
i.e., the point A, the corner of the
rectangle in 1st quadrant, is
(2, 1). Again the ellipse circumscribing the rectangle passess 1 l Bisector
- 4
through the point (4, 0), so its equation is = - 3 = ( k - 1)
x2 y 2 k - 1 P(1, 4) Q(k , 3)
16 b2
+ = 1
The midpoint = æ k + 1 , 72 øö
è 2
A(2, 1) lies on the above ellipse
Þ 146 + 1 = 1 Þ 1 = 1 - 1 = 3 The equation to the bisector l is
b2 b2 4 4
Þ b2 = 4/3 æ y - 7ö = (k - 1) æè x - k + 1ö
è 2ø 2 ø
Thus the equation to the desired ellipse is
x 2 + 3 y 2 = 1 Þ x2 + 12y2 = 16 As x = 0, y = – 4 satisfies it, we have
16 4
23. (a) : The radical axis, which in the case of intersection of the æ -4 - 7ö = (k - 1) èæ 0 - k + 1ö Þ 15 k 2 -1
è 2ø 2 ø - =-
circles is the common chord, of the circles
2 2
S1 : x 2 + y 2 + 3x + 7y + 2p – 5 = 0 and Þ k 2 – 1 = 15 Þ k 2 = 16 \ k = ± 4.
S2 : x2 + y 2 + 2x + 2y – p2 = 0 is S1 – S2 = 0
Þ x + 5y + 2p – 5 + p2 = 0 ...(i) 28. (b, c) : Equation of normal at P(x, y) is
If there is a circle passing through P, Q and (1, 1) it’s necessary Y - y = - ddyx ( X - x ) ÞG º æ x + y × ddxy ,0 ÷øö
çè
and sufficient that (1, 1) doesn’t lie on PQ,
i.e., 1 + 5 + 2p – 5 + p 2 ¹ 0 x + y dy = | 2 x | Þ y dy = x or y dy = - 3 x
dx dx dx
Þ p 2 + 2p + 1 ¹ 0 Þ (p + 1) 2 ¹ 0 \ p ¹ –1
ydy = xdx or ydy = –3xdx
Thus for all values of p except ‘–1’ there is a circle passing through
P, Q and (1, 1). After integrating, we get
24. (b) : Obviously the major axis is along the xaxis y2 = x2 +c or y 2 = - 3 x 2 + c
The distance between the focus and the corresponding directrix 2 2 2 2
Þ x 2 – y 2 = –2c
= a = 4 or 3x2 + y2 = 2c Þ x2 – y 2 = c 1 or 3x2 + y2 = c 2.
- ae
e
76 JEE MAIN CHAPTERWISE EXPLORER
29. (d) : Equation of circle (x – h)2 + (y – k) 2 = k2 Þ h2 + k 2 = 9 Þ x2 + y 2 = 94
It is passing through (–1, 1) then 4
(–1 – h) 2 + (1 – k) 2 = k 2 Þ h 2 + 2h – 2k + 2 = 0,
D ³ 0 37. (d) : Let OA = r
2k -1 ³ 0 Þ k ³ 21 . Given area = 49 p
30. (a) : Sum of the slopes = - co efficient of xy P A
co efficient of y 2 O 3x – 4y = 7
2x – 3y = 5
\ Sum of slopes = - (1- m2 ) = 0 B Q
m
Þ pr2 = 49 p
Þ m = ± 1. r = 7
Second method Angle bisector R (3, 3 3) Point of intersection of AB and PQ is (1, –1)
Equation of P(–1, 0) 120° \ equation of circle is (x – 1) 2 + (y + 1) 2 = 7 2
bisectors Þ x 2 + y2 – 2x + 2y – 47 = 0
of lines 60°
xy = 0 are y = ± x Q(0, 0)
Put y = ± x in 38. (a) : Let x 2 + y 2 =1 (a > b )
my 2 + (1 – m2 )xy – mx 2 = 0, we get a2 b2
Given 2b = 8 ... (i)
... (ii)
(1 – m 2 )x2 = 0 Þ m = ± 1. and 2ae = 6
31. (c) : Slope of the required angle bisector is tan120° = - 3 (o, b )
Hence equation of the angle bisector is
(o, o ) (a, o )
y = - 3( x - 0) A¢ F¢ F A
Þ 3x + y = 0 A(1, k )
32. (a) : 12 ´ | k -1 | ´1 = 1 By (i) and (ii) we have b = 4
k = –1, 3. ae 3
B(1, 1) C(2, 1) Þ b 2 = 16 e 2
a2 9
33. (b) : Let P is the required point, then P lies on directrix Þ 1 - e2 = 196 e2 (Q b2 = a2 (1 – e 2 ) as a > b)
x = –2 of y2 = 8x
Hence P º (-2 ,0). Þ e = 3
5
34. (b) : Q b 2 = a 2 (e 2 – 1) Þ sin2 a = cos2 a (e2 – 1) Þ tan2 a 39. (a) : Let h, k be the locus of the vertex of family of parabola
+ 1 = e2 Þ e2 = sec2 a
Vertices º (±cosa, 0) y = a3x2 + a2 x - 2 a
Coordinate of focii are (±ae,0) º (±1,0) 3 2
Þ if a varies then the abscissa of foci remain constant.
\ k = a3h2 + a 2 h = 2a
3 2
35. (c) : Given lines are y = 2x ( x > 0) Þ 3k = h 2 + 32a h - 6
and y = 3x (x > 0) using (a, a 2) in these lines a3 a 2
a2 - a > 0 ... (i) Þ 3æ k + 35a ö = æ h + 3 ö 2
2 ... (ii) a3 è 16 ø è 4 a ø
and a 2 – 3a < 0 { } i.e., 3 35a
a3 43a , y 16
Solving (i) and (ii) we get 12 < a < 3 x2 = y, where x = h + = k +
36. (d) : Let AB is chord of circle and M(h, k) be mid point of Þ vertex is çæè -3 , -1365 a ÷öø2
AB Ð AOM = 60° 4a
Now OA = OB = 3 and
OM ^ AB (By properties of circle) \ hk = çæè -3 ÷öø çæè -1365 a ÷øö
4a
Now OA = h2 + k 2 , OM = r cos q O (o, o ) Þ hk = 105
h2 + k 2 = 3cos 60 ° M(h, k) 64
h2 k 2 3 A B Þ xy = 16045 (taking h = x, k = y)
2
+ =
Two Dimensional Geometry 77
40. (c) : Now the equation of line which meet the xaxis and y Þ b 2 = a2 e 2 Þ a2 – a2 e 2 = a 2e 2
axis y (0, b ) B (a, b ) æ Q e 2 b 2 ö
F F¢ ç
= 1 - a2 ÷
ø
B è
Mid point (a, b) = M (3, 4)
o x Þ 1 – e 2 = e 2, 2e 2 = 1 , e = ± 1 .
A(a , 0) 2
at A(a, 0) B (0, b) is given by 44. (c) : Let locus of the centre of circle be (a, b).
x + y = 1 If C1 , C2 are centres of the circles with radii r1 , r 2 respectively
a b then (C 1C 2 ) 2 = r1 2 + r 22
Þ a 2 + b 2 = p2 + (a – a)2
where a = 2a = 6
and b = 2b = 8
6x + y + (b – b)2 (a, b)
8
\ required equation be = y Þ p 2 + a2 + b 2 – 2aa – 2bb
Þ 4x + 3y = 24 = 0 (a, b)
Þ 2ax + 2by – (a2 + b2 + p 2) (0, 0)
41. (c) :
ax2 + 2(a + b)xy + by2 = 0 (*) = 0.
Let q be the angle between the lines represent by * 45. (c) : Let locus of centre be a, b then according to given, if
r1 , r2 are radii of circles then
\ tan q = 2 h2 - ab
a + b y
2 (a + b) 2 - ab C (0, 5)
tan q = B (a, b)
a+b q Case (i) (0, 0) x
A
{3 area O¼AB = Area of D¼BC} D Internal touch. This case does not exist as centre
of circle is (0, 3) and radius is 2.
Now q = 45
{Q area of one sector = 3 time the C1 C 2 = r 2 ± r 1
area of another sector} Þ (a – 0)2 + (b – 3)2
2 (a + b) 2 - ab = |b ± 2|
\ tan45° =
Þ a 2 + b 2 – 6b + 9 = b 2 + 4 + 4b
a+b
and a 2 + b 2 – 6b + 9 = b 2 – 4b + 4
Þ 3a 2 + 2ab + 3b2 = 0.
Þ a 2 – 10b + 5 = 0 and x 2 = 2b + 5
42. (c) : Given y = ax + b and x 2 - y 2 = 1
a2 b2 Þ x 2 = 10y – 5 and x 2 = 2y – 5
\ b2 x2 - a2 y2 = a2b2
Both are parabolas but x2 = 2y – 5 does not exist.
Þ b2 x2 - a2 (ax + b) 2 = a2b2
46. (a) : As the line passes through P and Q which are the point
by using y = aa + b of intersection of two circles. It means given line is the equation
of common chord and the equation of common chord of two
Þ x2 ( b 2 – a 2a 2) – 2a2 abx + (–b 2a 2 – a2 b 2) = 0 ....(*) intersecting circle is
S1 – S2 = 0
Now the line y = ax + b will be tangent to circle if both roots = 5ax + (c – d)y + a + 1 = 0.
Now 5ax + (c – d)y + a + 1 = 0 and
if (*) are equal 5x + by – a = 0 represent same equation.
\ keeping D = 0 in (*) we have \ 5a = c - d = a + 1
4a2a4b2 = 4(b2 - a2a2 )(-b2a2 - a2b2 )
Þ a2a2b2 = (b2 - a2a2 )(-b2 - b2 )
Þ a2a2b2 = - b2b2 + b2a2a2 - b4 + a2a 2b2
Þ a2a2b2 = b2 (b2 + b2 ) Þ a2a2 = b2 + b 2
Þ a2 x2 - y2 = b2 .
43. (b) : F ¢(–ae, 0), F(ae, 0) 5 b -a
Slope of BF = b = m1 (say) Þ a2 + a + 1 = 0 and c - d +1 = - 1
ae ba
Slope of BF ¢ = b = m2 (say) ( ) Þ a + 1 2 3 = 0 and –(c – d + b) = b/a
- ae 2
+ 4
Now m 1 × m 2 = –1 Þ b ´ b = -1 d – b – c = +b/a has no solution. \ No value of a exist.
ae -ae
78 JEE MAIN CHAPTERWISE EXPLORER
47. (d) : \ x2 = 2(– 1) – 1 = – 3 Q(x2, y2 ) 52. (c) : The point of intersection of parabola’s y2 = 4ax and
x 2 = 4ay are A(0, 0), B(4a, 4a)
y2 = 2 × 2 – 1 = 3 M(–1, 2)
as the line 2bx + 3cy + 4d = 0 passes through these points
x 3 = 3 × 2 – 1 = 5 G \ d = 0 and 2b(4a) + 3c(4a) = 0
Þ 2b + 3c = 0 Þ (2b + 3c) 2 + d2 = 0
y3 = 2 × 2 – 1 = 3
P(1, 1) M2 (3, 2) R(x3, y3 )
\ Centroid G = æ x1 + x2 + x3 , y1 + y2 + y3 ö 53. (c) : Given circle x2 + y 2 – 2x = 0 ... (1)
çè 3 3 ÷ø
and line be y = x ... (2)
Solving (1) and (2) we get
( ) ( ) = 1- 3 + 5 , 1+ 3 + 3 = 1, 7 . x = 0, 1 \ y = 0, 1 y = x B
C
33 3
A
48. (d) : Let us take the two set of values of \ A(0, 0), B(1, 1) and equation of circle in the diameter form
a = 1, b = 1/2, c = 1/3
and a = 1/2, b = 1/3, c = 1/4 is (x – 0) (x – 1) + (y – 0) (y – 1) = 0
Putting these value in the given equation we get
x + 2y + 3 = 0 and 2x + 3y + 4 = 0 Þ x 2 + y 2 – (x + y) = 0
Solving the equations of (*) we have x = 1, y = –2
(1, –2) is required point on the line. ...(*) 54. (c) : As per given condition centre of the circle is the point
of intersection of the 2x + 3y + 1 = 0 and 3x – y – 4 = 0
49. (b) : Intercepts made by the lines with coordinate axis is
\ centre is (1, –1)
(–3b/a , 0), (0, –3/2) y Also circumference of the circle be given 2pr = 10p
\ r = 5
and (0, –3/2) \ Required equation of circle is
(x – 1)2 + (y + 1) 2 = 5 2
and (3a/b, 0) and (–3b/a, 0) (–3a/b, 0) or x 2 + y2 – 2x + 2y – 23 = 0
Common intercept is (0, –3/2). (0, –3/2)
55. (c) : Equation of circle AB as diameter is given by
50. (d) : Let M(x¢, y¢) be point of locus mid point of PQ. (x – p) (x – a) + (y – q) (y – b) = 0
Þ x ¢ + 1 = h, y ¢ + 0 = k , P(1, 0) Þ x 2 + y 2 – x(p + a) – y (q + b) + pa + qb = 0 ...(1)
2 2
M(h, k ) Now (1) touches axis of x so put y = 0 in (1) we have
\ x¢ = 2h – 1, y¢ = 2k Q(x¢, y ¢ )
x2 – x(p + a) + pa + qb = 0 ...(2)
Now (x¢, y¢) lies on y2 = 8x
Þ (2k) 2 = 8(2h – 1) and D = 0 in equation (2) B(a, b)
\ (p + a)2 = 4[pa + qb]
Þ y 2 = 2(2x – 1) Þ y 2 – 4x + 2 = 0.
51. (b) : Equation of directrix x = 4 which is parallel to yaxis Þ (p – a) 2 = 4qb A(p, q)
Now a ® x, b ® y
so axis of the ellipse is xaxis. Let equation of ellipse be xaxis
x2 + y 2 = 1 (a > b) \ (p – x)2 = 4q(y) which required locus of one end point of
a2 b2
the diameter.
Again e = 1/2 and e 2 = 1 - b 2 56. (b) : Let the equation of circle cuts orthogonally the circle x 2
a2
æ b ö2 + y 2 = 4 is
Þ çè a ÷ø = 1 – 1/4 = 3/4 ...(*) x 2 + y2 + 2gx + 2fy + c = 0 ...(i)
Also the equation of one directrix is x = 4 \ 2g 1g 2 + 2f 1 f 2 = c 1 c 2 (where (–g, –f) are point of locus)
Þ c = – 4
a Again circle (i) passes through (a, b), so
\ equation of directrix x = e
a 2 + b 2 + 2ga + 2fb + 4 = 0
a
\ 4 = e Now replacing g, f by x, y respectively
Þ a = 2 (³ e = 1/2) \ 2ax + 2by – (a 2 + b2 + 4) = 0
Further b2 = a2 ´ 3 by (*) 57. (d) : The equation ax 2 + 2hxy + by2 = 0
4
= (y – m 1x ) (y – m2 x)
b 2 = 4 ´ 3 = 3 Þ m 1 + m2 = - 2h = 1 ...(*)
4 b 4c
x2 y 2 m 1m 2 = 3
Hence equation of ellipse is a2 b2 c
+ = 1 2
x2 y 2 and 3x + 4y = 0 Þ m 1 = –3/4
+
Þ 4 3 = 1 or 3x 2 + 4y 2 = 12 \ m2 = - 2
c
Two Dimensional Geometry 79
Now by (*) we have Þ y = 9/2 \ x = y2 = 81 = 9
18 72 8
- æ 3+ 2 ö = 1 9
çè 4 c ÷ø 4c
So the required point is x = 8 , y = 9/2
Þ - 3 = 1 + 2 Þ - 3 = 1 + 8
4 4c c 4 4c 4c
63. (d) : Let a, b is the point of locus, equidistant from (a 1 , b 1)
3 9 and (a2 , b2 ) is given by
Þ - = 4c \ c = –3 (a – a1 ) 2 + (b – b1 ) 2 = (a – a 2) 2 + (b – b 2) 2
4 Þ a 1 2 + b1 2 – 2a1 a – 2b1 b – a 22 – b 22 + 2a2 a + 2b 2b = 0
Þ 2(a2 – a1 )a + 2(b 2 – b 1 )b + a1 2 + b 12 – b 2 2 – a2 2 = 0
58. (a) : If m 1 and m 2 are slope of the lines then by given condition
m 1 + m 2 = 4m 1 m 2
Þ - 2 c = - 4 Þ (a 2 – a1 )x + (b2 – b1 ) y +
7 7
Þ c = 2 1 (a1 2 + b1 2 – a2 2 – b2 2 ) = 0
2
By using ax2 + 2hxy + by2 = 0
1
- 2h a Þ c = - [a1 2 + b1 2 – a2 2 – b2 2 ]
Þ m1 + m 2 = b and m 1m 2 = b 2
59. (d) : Given OA + OB = –1 64. (a) : Let (h, k) be the coordinate of centroid
i.e. a + b = –1 y \ h = a cos t + b sin t + 1 ,
\ equation of the line be (4, 3)
3
B (b, 0)
x - y = 1 a sin t - b cos t + 0
a 1 + a A(a, 0) O k = 3
4 - 3 x Þ 3h – 1 = a cos t + b sin t ...(i)
a 1 + a
Þ = 1 3 k = a sin t – b cos t ...(ii)
Þ a = ± 2 (as a = 2 b = –3 By squaring (i) and (ii) then adding we get
and a = –2 b = 1) (3h – 1)2 + (3k)2 = a2 (cos2 t + sin2 t) + b 2 (cos 2 t + sin2 t )
x y x y Replacing (h, k) by (x, y) we get choice (a) is correct.
2 3 = -2 1 = 1
so equation are - 1 and + 65. (c) : Given equations are
x 2 – 2qxy – y2 = 0
60. (c) : Let locus of point C(h, k) and centroid (a, b) x2 – 2pxy – y 2 = 0 ...(1)
...(2)
As (a, b) lies on 2x + 3y = 1 \ 2a + 3b = 1
Joint equation of angle bisector of the line (i) and (ii) are same
Now centroid of ABC is
x2 - y2 xy
æ 2 + (-2) + h -3 + 1 + k ö æ h k - 2 ö \ =
çè 3 3 ÷ø or çè 3 , 3 ÷ø -q
1 + 1
æ h ö 3(k - 2) Þ qx 2 + 2xy – qy2 = 0 ...(3)
\ 2 çè 3 ÷ø + 3 = 1 Now (2) and (3) are same, taking ratio of their coefficients
1 - p
Þ 2h + 3k = 9 Þ 2x + 3y = 9
\ =
q 1
61. (c) : The equation of normal at q is Þ pq = –1
1 66. (c) : According to the problem square lies above xaxis
y – y 1 = - (x – x 1) Now equation of AB using two point form. We get
dy y – y1 = m(x – x 1)
(y – a sin a) = - a (cos a - sin a) [x – a cos a]
dx a (cos a + sin a )
sin q Þ y(cos a + sin a) + x(cos a – sin a)
Þ y – a sin q = cos q (x – a(1 – cos q)) = a sin a(cos a + sin a) + a cos a (cos a – sin a)
= a(sin2 a + cos2 a)
which passes through (a, 0)
= a(1)
dy dx
62. (d) : Given y2 = 18x and dt = 2 dt 67. (b) : Coordinate of centre may be (1, –1) or
(–1, 1) but 1, –1 satisfies the given equations of diameter, so
dy dx choices (a) and (d) are out of court.
\ 2y dt = 18 dt Again pR2 = 154, R2 = 49 \ R = 7
dy 18 dy
Þ 2y dt = 2 dt
80 JEE MAIN CHAPTERWISE EXPLORER
\ Required equation of circle be Again product of the slope of AC and AB
(x – 1) 2 + (y + 1) 2 = 49 1
Þ x2 + y 2 – 2x + 2y = 47
= × (–5) = –1
68. (d) : (x – 1) 2 + (y – 3)2 = r 2 \ C1 ( 1, 3) and r1 = t 1 = r 5
(x – 4) 2 + 1(y + 1)2 = 9 \ C2 ( 4, –1) and r 2 =t2 = 3
Þ AC b AB
so C1 C2 = (4 - 1)2 + (3 + 1)2 = 5 Þ right at A
Now for intersecting circles
r1 + r 2 > C1 C 2 and |r 1 – r 2| < C 1C 2 72. (c) : Given median of the equilateral triangle is 3a.
\ Þ r + 3 > 5 and |r – 3| < 5
Þ r > 2 and –5 < r – 3 < 5 In D LMD, (LM) 2 = (LD)2 + (MD) 2
Þ r > 2 and –2 < r < 8
Þ r Î (2, 8) (LM) 2 = 9a 2 + æ LM ö2
çè 2 ÷ø
Þ 3 (LM) 2 = 9a2 \ (LM) 2 = 12a 2
4
69. (d) : Since the normal at (bt 12 , 2bt1 ) , on parabola y2 = 4bx Again in triangle OMD, (OM) 2 = (OD) 2 + (MD) 2
meet the parabola again at (bt2 2 , 2bt 2 )
\ t 1x + y = 2bt1 + bt1 3 passes through (bt 22 , 2bt2 ) R 2 = (3a – R) 2 + æ LM ö2 L
çè 2 ÷ø M R
Þ R 2 = 9a 2 + R 2 – 6aR + 3a 2
Þ t 1b t 2 2 + 2bt2 = 2bt1 + bt1 3 O
Þ 6aR = 12a2
D
Þ t 1( t 2 2 – t1 2) = 2(t 1 – t2 ) R = 2a N
Þ t1 ( t2 + t 1 ) = –2 So equation of circle be
(x – 0) 2 + (y – 0)2 = R 2 = (2a) 2
Þ t 2 + t1 = - 2 Þ x 2 + y 2 = 4a 2
2
t1
Þ t 2 = - – t 1 73. (b) : Let equation of circle be x2 + y2 + 2gx + 2fy + c = 0.
t1 x2 As it passes through (0, 0) so c = 0
y 2
70. (b) : Eccentricity for + = 1 and as it passes (1, 0) so –g = 1 .
a2 b2 2
is b2 = a2 ( 1 –e2 )
Now x2 + y2 + 2gx + 2fy = 0 and x2 + y2 = 9 touches each other.
x2 - y2
and eccentricity for 144 81 = 1 is \ equation of common tangent is 2gx + 2fy –9 = 0 and distance
from the centre of circle x2 + y2 = 9 to the common tangent is
25 25 equal to the radius of the circle x2 + y2 = 9
ì a12 + b1 2 0 + 0 - 9
ïï e1 = a1 2
í \ 4g 2 + 4 f 2 = 3 3 3
ïîï\ e1 = 1 + 81 = 15 Þ 32 = 4(g2 + f 2 ) 9 (0,0)
144 12
= 4 çèæ 1 2 ö x 2 + y2 = 9 x2 + y2 + 2gx + 2fy = 0
12 ´ 15 4 + f ÷ø Common tangent
Again foci = a 1 e 1 = 5 12 = 3
2gx + 2fy = 9
\ focus of hyperbola is (3, 0) = (ae, 0) 9 = 1 + 4f 2
so focus of ellipse (ae, 0) = (4e, 0) \ f 2 = 2
As their foci are same \ 4e = 3 \ e = 3/4 f = ± 2
\ –f = ± 2
\ e2 = 1 – æ b ö2 b 2
çè a ÷ø = 1 – 16
\ Centre of the required circle be æ 1 , 2 ö , æ 1 - 2 ÷øö
b 2 çè 2 ÷ø çè ,
or = 1 – e 2 = 1 – 9
2
16 16 74. (d) : \ (h, k) is æ p , p ö
çè cos sin a ÷ø
Þ b 2 = 7 2 a 2 y
71. (a) : AB = 26 , AC = 26 p 0, p
\ cos a = , sin a
Y
B(3, 5) 2h
p Midpoint (h, k)
sin a =
2k (0,0) x
A Þ sin 2a + cos 2a æ p , 0 ÷ö
X ç cos a ø
O è
(4, 0)
C(–1, –1) p2 p 2 1 1 4
\ ABC is isosceles = 4h2 + 4k 2 = x2 + = p 2
y 2
Two Dimensional Geometry 81
75. (a) : Using fact : Pair of lines Ax 2 + 2hxy + By2 = 0 are b to abc + 2fgh – af 2 – bg2 – ch 2 = 0 ...(4)
Þ (abc – af 2 ) + (fgh – bg2 ) + fgh – ch 2 = 0
each other if A + B = 0 Þ 0 + 0 + fgh – ch 2 = 0
Þ 3a + a 2 – 2 = 0 \ ch 2 = fgh
Þ a2 + 3a – 2 = 0
Now adding (2) and (4)
Þ There exist two value of a as D > 0 2fgh = ch 2 + bg2
\ a = -3 ± 17
2
78. (c) : Equation of chord y = mx + 1
76. (a) : Let (a, b) is the centre of the circle whose radius is 3. Equation of circle x 2 + y2 = 1
\ Equation of such circle be Joint equation of the curve through the intersection of line and
circle be given by x2 + y 2 = (y – mx) 2,
(x – a)2 + (y – b)2 = 32 Þ x2 ( 1 – m2 ) + 2mxy = 0
Þ a 2 + b 2 – 2ax – 2by + 25 = 9
Þ x 2 + y2 – 2x2 – 2y 2 + 25 = 9
Þ x 2 + y2 = 25 – 9 h2 - ìï a = 1 - m 2
where í
Þ x 2 + y 2 = 16 and x 2 + y 2 = 25 Now tan q = ± 2 ab
Þ 4 £ x2 + y2 £ 64 a + b ïî h = m,b = 0
77. (a) : As s = ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 represent a 2 m 2 - 0
pair of line tan 45° = ± 1 - m2
Þ 1(1 – m 2 ) = ± 2m
a h g Þ m2 ± 2m – 1 = 0
\ h b f = 0
Þ m = ± 1 ± 2
g f c
...(1) Þ m = 1 ± 2 and –1 ± 2
or abc + 2fgh – af 2 – bg2 – ch 2 = 0 79. (b) : Let common tangent to the curves be
Now say point of intersection on y axis be (0, y1 ) and point of y = mx + c ...(1)
intersection of pair of line be obtained by solving the equations
a
¶s ¶s = mx + m
¶ x = 0 = ¶ y and \ y 2 = 8ax = 4(2a)x
\ Equation of tangent to parabola
¶s
\ ¶ x = 0 Þ ax + by + g = 0 Þ 2a
y = mx +
¶s ...(2)
m
and = 0 Þ hx + by + f = 0 Þ which is also tangent to the circle
¶ y
x 2 + y2 = 2a2 = ( 2a ) 2
ì hy1 + g = 0 (*)
îí by1 + f = 0 Now Distance from (0, 0) to the tangent line = Radius of circle
On compairing the equation given in (*) we get \ 2 a = ± 2a ´ 1 (0 0)
m 1 + m2
bg = fh and bg 2 = fgh ...(2)
Again ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 Þ m2 ( 1 + m 2 ) – 2 = 0
meet at yaxis \ x = 0 Þ (m 2 – 1) (m 2 + 2) = 0 y = mx + 2a
Þ by2 + 2fy + c = 0 whose roots must be equal Þ m = ± 1 m
\ f 2 = bc
...(3) 2a
af 2 = abc Required equation of tangent y = mx + m
= ± (x + 2a)
Now using (2) and (3) in equation (1) we have
82 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER THREE DIMENSIONAL
GEOMETRY
13
1. If the lines x-2 = y-3 = z - 4 and x -1 = y-4 = z - 5 are (a) 2/5 (b) 5/3 (2011)
1 1 - k k 2 1 (c) 2/3 (d) 3/2
coplanar, then k can have 7. A line AB in threedimensional space makes angles 45° and
(a) exactly three values (b) any value 120° with the positive xaxis and the positive yaxis respectively.
(c) exactly one value (d) exactly two values If AB makes an acute angle q with the positive zaxis, then
(2013) q equals
2. Distance between two parallel planes 2x + y + 2z = 8 and (a) 30º (b) 45º
4x + 2y + 4z + 5 = 0 is (c) 60º (d) 75º. (2010)
5 7 9 (d) 3 8. Statement1 : The point A(3, 1, 6) is the mirror image of the
(a) 2 (b) 2 (c) 2 2 point B (1, 3, 4) in the plane x – y + z = 5.
(2013) Statement2 : The plane x – y + z = 5 bisects the line segment
3. An equation of a plane parallel to the plane joining A (3, 1, 6) and B(1, 3, 4).
x – 2y + 2z – 5 = 0 and at a unit distance from the origin is (a) Statement 1 is true, Statement2 is true; Statement2 is
(a) x – 2y + 2z – 1 = 0 (b) x – 2y + 2z + 5 = 0
(c) x – 2y + 2z – 3 = 0 (d) x – 2y + 2z + 1 = 0 a correct explanation of Statement 1.
(2012) (b) Statement1 is true, Statement2 is true; Statement 2 is not
a correct explanation for Statement1.
4. If the lines x - 1 = y + 1 = z - 1 and x - 3 = y-k = z intersect, (c) Statement1 is true, Statement2 is false.
2 3 4 1 2 1 (d) Statement1 is false, Statement2 is true.
(2010)
then k is equal to
(a) 9/2 (b) 0 (c) – 1 (d) 2/9 (2012) 9. Let the x - 2 y - 1 z + 2
3 = -5 = 2 line lie in the plane
5. Statement1 : The point A(1, 0, 7) is the mirror image of the
point B(1, 6, 3) in the line x + 3y – az + b = 0. Then (a, b) equals
x = y - 1 = z - 2 (a) (–6, 7) (b) (5, –15)
12 3 (c) (–5, 5) (d) (6, –17) (2009)
Statement2 : The line : x = y - 1 = z - 2 bisects the line 10. The projections of a vector on the three coordinate axis are
1 23 6, –3, 2 respectively. The direction cosines of the vector are
segment joining A(1, 0, 7) and B(1, 6, 3). (a) 6 , -3 , 2 (b) 6 , -3 , 2
(a) Statement1 is true, Statement2 is false. 5 5 5 7 7 7
(b) Statement1 is false, Statement2 is true.
(c) Statement1 is true, Statement2 is true; Statement2 is a (c) -6 -3 2 (d) 6, –3, 2 (2009)
correct explanation for Statement1. 7, 7 , 7
(d) Statement1 is true, Statement2 is true; Statement2 is 11. If the straight lines
not a correct explanation for Statement1. x -1 = y - 2 = z - 3 and x - 2 = y - 3 = z - 1
(2011) k23 3 k 2
6. If the angle between the line x = y -1 = z - 3 and the plane intersect at a point, then the integer k is equal to
2 l
(a) – 2 (b) – 5
x + 2 y + 3z = 4 is cos -1 èæç 5 ö
14 ÷ø then l equals (c) 5 (d) 2 (2008)
Three Dimensional Geometry 83
12. The line passing through the points (5, 1, a) and (3, b, 1) (a) - 3 (b) 5 (c) - 4 (d) 3 .
5 3 3 4
æ 17 , -213 öø . Then
crosses the yzplane at the point è 0, 2 (2005)
22. The distance between the line
(a) a = 8, b = 2 (b) a = 2, b = 8 r
r = 2iˆ - 2 ˆj + 3kˆ + l(iˆ - ˆj + 4kˆ ) and the plane
(c) a = 4, b = 6 (d) a = 6, b = 4 (2008) rr × (iˆ + 5 ˆj + kˆ ) = 5 is
13. Let L be the line of intersection of the planes 10 10
(a) 3 3 9
2x + 3y + z = 1 and x + 3y + 2z = 2. If L makes an angle a (b)
with the positive xaxis, then cos a equals 10 3 .
3 10
(a) 1 1 1 (d) 12 . (c) (d) (2005)
(b) 2 (c) 3
23. If the plane 2ax – 3ay + 4az + 6 = 0 passes through the
r (2007)
Let ar = iˆ + ˆj + kˆ, b
14. = iˆ - ˆj + 2 kˆ and midpoint of the line joining the centres of the
crr = xiˆ + (x - 2) ˆj - kˆ . If the vectors cr lies in the plane of ar and spheres x 2 + y 2 + z 2 + 6x – 8y – 2z = 13 and
x2 + y2 + z 2 – 10x + 4y – 2z = 8 then a equals
b , then x equals (a) 1 (b) –1 (c) 2 (d) –2. (2005)
(a) – 4 (b) –2 (c) 0 (d) 1. 24. The intersection of the spheres
(2007) x2 + y2 + z2 + 7x – 2y – z = 13 and
15. If (2, 3, 5) is one end of a diameter of the sphere x2 + y 2 + z2 – 3x + 3y + 4z = 8 is the same as the intersection
x 2 + y 2 + z 2 – 6x – 12y – 2z + 20 = 0, then the coordinates of one of the sphere and the plane
of the other end of the diameter are (a) x – y – 2z = 1 (b) x – 2y – z = 1
(a) (4, 3, 5) (b) (4, 3, –3) (c) x – y – z = 1 (d) 2x – y – z = 1. (2004)
(c) (4, 9, –3) (d) (4, –3, 3). (2007) 25. A line with direction cosines proportional to
2, 1, 2 meets each of the lines x = y + a = z and
16. If a line makes an angle of p/4 with the positive directions x + a = 2y = 2z. The coordinates of each of the points of
intersection are given by
of each of xaxis and yaxis, then the angle that the line (a) (3a, 2a, 3a), (a, a, 2a) (b) (3a, 2a, 3a), (a, a, a)
(c) (3a, 3a, 3a), (a, a, a) (d) (2a, 3a, 3a), (2a, a, a).
makes with the positive direction of the zaxis is (2004)
(a) p (b) p (c) p (d) p .
4 2 6 3
(2007)
17. The image of the point (–1, 3, 4) in the 3 plane 26. Distance between two parallel planes
x – 2y = 0 is
2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is
æ - 17 , - 19 , 4 ÷öø (a) 7/2 (b) 5/2
çè 3 3
(a) (b) (15, 11, 4) (c) 3/2 (d) 9/2 (2004)
(c) æ - 17 , - 19 , 1 ÷øö (d) çèæ 9 , - 13 , 4 ÷öø . (2006) 27. A line makes the same angle q, with each of the x and z axis.
çè 3 3 5 5
If the angle b, which it makes with yaxis, is such that
18. The two lines x = ay + b, z = cy + d and sin2 b = 3sin 2q, then cos2 q equals
x = a¢y + b¢, z = c¢y + d¢ are perpendicular to each other if (a) 3/5 (b) 1/5
(a) aa¢ + cc¢ = –1 (b) aa¢ + cc¢ = 1 (c) 2/3 (d) 2/5. (2004)
a + c = -1 a + c = 1 . 28. If the straight lines x = 1 + s, y = –3 – ls, z = 1 + ls and
a¢ c¢ a¢ c¢
(c) (d) (2006) t ,
2
x = y = 1 + t, z = 2 – t, with parameters s and t respectively,
19. The angle between the lines 2x = 3y = –z and 6x = –y = –4z are coplanar, then l equals
is (a) –1/2 (b) –1 (c) –2 (d) 0. (2004)
(a) 90º (b) 0º (c) 30º (d) 45º. (2005) y - 3 y - 4
1 2
20. The plane x + 2y – z = 4 cuts the sphere 29. The lines x-2 = = z - 4 and x -1 = = z - 5
1 - k k 1
x 2 + y2 + z2 – x + z – 2 = 0 in a circle of radius
are coplanar if
(a) 1 (b) 3 (c) 2 (d) 2. (2005) (a) k = 1 or –1 (b) k = 0 or –3
x + 1 y 2-1 = z - 2 (c) k = 3 or –3 (d) k = 0 or –1. (2003)
1 2
21. If the angle q between the line = and the 30. The two lines x = ay + b, z = cy + d and
x = a ¢y + b ¢, z = c ¢y + d will be perpendicular, if and only
plane 2x - y + lz + 4 = 0 is such that sin q = 1 , the value if
3
of l is
84 JEE MAIN CHAPTERWISE EXPLORER
(a) aa¢ + bb¢ + cc¢ = 0 34. The shortest distance from the plane
(b) (a + a¢) (b + b¢) + (c + c ¢) = 0
(c) aa¢ + cc¢ + 1 = 0 12x + 4y + 3z = 327 to the sphere
(d) aa¢ + bb¢ + cc¢ + 1 = 0.
x 2 + y 2 + z 2 + 4x – 2y – 6z = 155 is
(2003) (a) 11 34 (b) 13
(c) 39 (d) 26.
a a2 1 + a 3 (2003)
31. If b b2 1 + b 3
= 0 and vectors (1, r , ar 2 ) , (1, r r and 35. Two systems of rectangular axes have the same origin. If a
a b, b 2 ) plane cuts them at distance a, b, c and a ¢, b ¢, c ¢ from the
origin, then
c c2 1 + c3
(1, r cr 2 ) are noncoplanar, then the product abc equals (a) 1 + 1 - 1 + 1 + 1 - 1 = 0
c, a2 b2 c2 a¢2 b¢2 c¢2
(a) –1 (b) 1 (c) 0 (d) 2. (2003) 1 1 1 1 1 1
a2 b2 c2 a¢2 b¢2 c¢ 2
32. A tetrahedron has vertices at O (0, 0, 0), (b) - - + - - = 0
A (1, 2, 1), B (2, 1, 3) and C (–1, 1, 2). Then the angle between 1 1 1 1 1 1
a2 b2 c2 a¢2 b¢2 c¢ 2
the faces OAB and ABC will be (c) + + - - - = 0
(a) cos –1( 17/31) (b) 30°
(c) 90° (d) cos– 1 (19/35). (2003) (d) 1 + 1 + 1 + 1 + 1 + 1 = 0. (2003)
a2 b2 c2 a¢2 b¢2 c¢2
33. The radius of the circle in which the sphere
x 2 + y 2 + z 2 + 2x – 2y – 4z – 19 = 0 is cut by the plane 36. The d.r. of normal to the plane through (1, 0, 0), (0, 1, 0)
which makes an angle p/4 with plane x + y = 3 are
x + 2y + 2z + 7 = 0 is
(a) 2 (b) 3 (a) 1, 2, 1 (b) 1, 1, 2
(c) 4 (d) 1. (2003) (c) 1, 1, 2 (d) 2, 1, 1. (2002)
Answer Key
1. (d) 2. (b) 3. (c) 4. (a) 5. (d) 6. (c)
7. (c) 8. (b) 9. (a) 10. (b) 11. (b) 12. (d)
13. (c) 14. (b) 15. (c) 16. (b) 17. (d) 18. (a)
19. (a) 20. (a) 21. (b) 22. (a) 23. (d) 24. (d)
25. (b) 26. (a) 27. (a) 28. (c) 29. (b) 30. (c)
31. (a) 32. (d) 33. (b) 34. (b) 35. (c) 36. (b)
Three Dimensional Geometry 85
1. (d) : For the lines to be coplanar Angle between line and plane (by definition)
1 -1 -1 = sin-1 æ 1×1+ 2×2 + l×3 ö = sin -1 æç 5 + 3 l ö
1 1 -k = 0 ç ÷ ÷
k2 1 è 1+ 4 + 9 1+ 4 + l2 ø è 14 5 + l 2 ø
So, (5 + 3l) 2 + 5 = 1 (sin2 q + cos2 q = 1)
14(5 + l 2 ) 14
Expanding, we get 1(1 + 2k) + 1(1 + k2 ) – 1(2 – k) = 0
Þ k 2 + 1 + 2k + 1 – 2 + k = 0 Þ (5 + 3l) 2 + 5 = 14
Þ k 2 + 3k = 0 Þ k(k + 3) = 0 \ k = 0, –3 5 + l2
So there are two values of k.
Þ (5 + 3l) 2 + 5(5 + l 2) = 14(5 + l 2 )
2. (b) : The planes are Þ 25 + 30l + 9l 2 + 25 + 5l 2 = 70 + 14l 2
Þ 30l + 50 = 70
4x + 2y + 4z = 16, 4x + 2y + 4z = –5 Þ 30l = 20 \ l = 2/3
Distance between planes = 16 - (-5) = 21 = 7 7. (c) : We have l = 1 , m = - 1
42 + 22 + 42 6 2 2 2
3. (c) : Equation of a plane parallel to As l 2 + m2 + n 2 = 1, we have n2 = 1 Þ n = 1
42
x – 2y + 2z – 5 = 0 and at a unit distance from origin is x –
We take positive values, so n = 1
2y + 2z + k = 0 2
Þ |k| = 1 Þ |k|= 3 Þ c os q = 1 . \ q = 60°.
3 2
\ x – 2y + 2z – 3 = 0 8. (b) : Let the image be (a, b, c)
or x – 2y + 2z + 3 = 0 Thus by image formula, we have
4. (a) : x - 1 = y + 1 = z - 1 = r1
2 3 4
and x - 3 = y - k z a-1 = b-3 = c - 4 = -2 æçè 1 - 3+ 4 - 5 ö
1 2 1 1 -1 1 3 ø÷
= = r2
or 2r 1 + 1 = r 2 + 3, 3r 1 – 1 = 2r 2 + k, 4r 1 + 1 = r 2 Þ a - 1 = b - 3 = c - 4 = 2
Þ 2r 1 – r 2 = 2, and 4r1 – r 2 = – 1 1 -1 1
– 2r 1 = 3 Þ r1 = - 3 and r2 = - 5 \ (a, b, c) = (3, 1, 6)
2
Again the midpoint of A(3, 1, 6) and B(1, 3, 4) is
\ - 9 - 1 = - 10 + k Þ k = 10 - 11 = 9 (2, 2, 5) & the equation of the plane is x – y + z =5.
2 22
As the point lies on the plane, so the plane bisects the segment
5. (d) : The direction ratios of the line segment joining
A(1, 0, 7) and B(1, 6, 3) is (0, 6, – 4). AB. But it does not explain statement1.
9. (a) : The line is x - 2 = y -1 = z + 2
3 -5 2
The direction ratios of the given line is (1, 2, 3).
As 1.0 + 6.2 – 4.3 = 0 we have the lines as perpendicular The direction ratios of the line are (3, –5, 2).
Also the midpoint of AB lies on the given line, so statement As the line lies in the plane x + 3y – az + b = 0,
1 and statement 2 are true but statement 2 is not a correct
explanation of statement 1. we have (3)(1) + (–5)(3) + 2(–a) = 0
Þ –12 – 2a = 0. \ a = –6
Statement ‘2’ holds even if the line is not perpendicular. This Again (2, 1, –2) lies on the plane
situation is possible.
Þ 2 + 3 + 2a + b = 0
6. (c) : x - 0 = y - 1 = z - 3 Þ b = –2a – 5 = 12 – 5 = 7
1 2 l
Hence (a, b) is (–6, 7).
x + 2y + 3z = 4
86 JEE MAIN CHAPTERWISE EXPLORER
uuur Then cosa = 3 = 1 .
10. (b) : Let the vector PQ be (x1 – x2, y1 – y2, z1 – z2) 9 + 9 + 9 3
we have x1 – x2 = 6 Second method
y1 – y2 = –3
z1 – z2 = 2 If direction cosines of L be l, m, n, then
Length of PQ 2l + 3m + n = 0, l + 3m + 2n = 0
PQ = (x1 - x2 )2 + ( y1 - y2 )2 + (z1 - z2 ) 2
After solving, we get, l = m = n
= 62 + 32 + 22 = 36 +uu9ur+ 4 = 7 3 - 3 3
The direction cosines of PQ are
\ l : m : n = 1 : - 1 : 1 Þ cosa = 1 .
3 33 3
14. (b) : ar = iˆ + ˆj + kˆ , r iˆ - ˆj + 2 kˆ and
[crar =brxiˆcr+ ] b=
x1 - x2 y1 - y2 z1 - z 2 (x - 2) ˆj - kˆ
PQ PQ PQ = 0
, ,
1 1 1
6 3 2 1 -1 2 = 0
i.e., 7 , - 7 , 7
x x - 2 -1
11. (b) : As the lines intersect, Þ 1(1 – 2x + 4) –1 (–1 – 2x) + 1 (x – 2 + x) = 0
we have
which on solving given 2k 2 + 5k – 25 = 0 Þ 5 – 2x + 1 + 2x + 2x – 2 = 0
Þ 2k 2 + 10k – 5k – 25 = 0
Þ 2k(k + 5) – 5(k + 5) = 0 Þ x = – 2.
Þ (2k – 5) (k + 5) = 0
15. (c) : Centre of sphere º (3, 6, 1)
Let the other end of diameter is (a, b, g)
\ k = - 5, 5 3= a + 2 Þ a =4 , 6 = b + 3 Þ b = 9
2 2 2
1 = g + 5 Þ g = -3.
2
12. (d) : The equation of the line passing through 16. (b) : Let required angle is q
(3, b, 1) and (5, 1, a) is
Q p p4 then
x - 5 y - 1 z - a l = cos 4 , m = cos n = cos q
2 = 1 - b = a - 1 = m (say)
We know that l 2 + m2 + n2 = 1
The line crosses the yz plane where x = 0, i.e. Þ cos2 p + cos2 p + cos2 q = 1
4 4
- 5 = 2 m \ m = - 25 Þ 1 + 12 + cos2 q = 1
2
1 = 17
Again y = m (1 - b) + 2 Þ cos2 q = 0 Þ q = p/2
Thus required angle is p/2.
Þ - 5 (1 - b) + 1 = 17 Þ - 5 (1 - b) = 15
2 2 2 2
17. (d) : Image of point (x¢, y¢, z¢) in
Þ (1 - b) = -3 \ b = 4 ax + by + cz + d = 0 is given by
Again z = m( a - 1) + a = - 13 x - x¢ = y - y¢ = z - z ¢ = -2(ax¢ + by¢ + cz¢ + d )
2 a b c a2 + b2 + c2
5 - 123 3 5 - 123 Þ x +1 = y - 3 = z - 4 = -2(-1 - 6)
2 2 2 1 -2 0 5
Þ - ( a - 1) + a = Þ - a + =
Þ - 3 a = -9 Þ a = 6 \ x = 9 , y = -51 3 , z = 4
2 5
iˆ ˆj kˆ 18. (a) : Fact : Two lines x - x1 = x - y1 = z - z1
13. (c) : Direction of the line, L = 2 3 1 a1 b1 c 1
1 3 2 and x - x2 = y - y2 = z - z2
= 3iˆ - 3 ˆj + 3k ˆ . a2 b2 c2
are ^ if a 1a 2 + b 1 b 2 + c 1c 2 = 0
Three Dimensional Geometry 87
Given lines can be written as x-b = y = z - d ...(i) 23. (d) : Centre of spheres are (–3, 4, 1) and (5, –2, 1)
a 1 c ...(ii)
M (1, 1, 1)
x - b¢ y z - d ¢
and a¢ = 1 = c¢ C1 (–3, 4, 1) C2 (5, –2, 1)
As lines are perpendicular using mid point in the equation
2ax – 3ay + 4az + 6 = 0
\ a a¢ + 1 + c c¢ = 0 Þ 2a – 3a + 4a + 6 = 0 Þ a = –2.
Þ a a¢ + c c¢ = –1
19. (a) : From given line 24. (d) : Equation of the plane of intersection of two spheres
S1 = 0 = S2 is given by S 1 – S2 = 0
x = y = z and x = y = z
3 2 -6 2 -12 -3 Þ 10x – 5y – 5z = 5
cos q = a1a2 + b1b2 + c1c2
a12 + b12 + c12 a22 + b22 + c22 Þ 2x – y – z = 1
cos q = 6 - 24 + 18 = 0 25. (b) : Given AB = x = y + a = z
1 1 1
32 + 22 + (-6)2 22 + (-12)2 + (-3)2
\ q = 90º. CD : x+a = y = z
2 1 1
20. (a) : Centre of sphere is 1/2, 0, –1/2 Let P º (r, r – a, r) and Q = (2l – a, l, l)
R = Radius of sphere is g 2 + f 2 + w2 - c Direction ratios of PQ are r – 2l + a, r – l – a, r – l
According to question direction ratios of
= 1 + 1 + 2 \ R = 5 PQ are (2, 1, 2)
44 2
d = ^ distance from centre to the plane is equal to \ r - 2l + a = r -l -a = r-l
2 1 2
(i) (ii ) (iii)
1 + 0 + 12 - 4 3 . (ii) and (iii) Þ r – l = 2a ...(1)
2 6
d = , d = (i) and (iii) Þ l = a r = 3a, l = a
12 + 22 + 12 \ p º (3a, 2a, 3a) and Q º (a, a, a).
\ Radius of the circle 26. (a) : Let x 1 , y 1, z 1 be any point on the plane
= Radius of sphere – perpendicular distance from centre of 2x + y + 2z – 8 = 0
sphere to plane
\ 2x 1 + y1 + 2z1 – 8 = 0
r nr1
( ) =æ 5 ö2 - 9 = 15 - 9 = 1. y + 2z = 8 y 1, a× a× = d1
çè 2 ø÷ 6 66 (x 1, z 1) r =
n2
2x + d2
21. (b) : Angle between the line and plane 90–q Normal of 4x + 2y + 4z + 5 = 0
is same as the angle between the line line plane
and normal to the plane q
\ cos(90 - q) = a1a2 + b1b2 + c1c2 Plane
a12 + b12 + c12 a22 + b22 + c22 2(2x + y + 2z - 8) + 21 21 7
==
\ d = 42 + 22 + 4 2 6 2
Þ 31 = (1´ 2 + 2 ´ (-1)) + 2 l 27. (a) : If a line makes the angle a, b, g with x, y, z axis respectively
12 + 22 + 22 22 + 12 + l
then
Þ l = 5 . l 2 + m 2 + n 2 = 1
3
r nr - Þ 2l 2 + m2 = 1 or 2n 2 + m2 = 1
22. (a) : d = a × d Þ 2 cos2 q = 1 – cos 2b (a = g = q)
n 2 cos 2q = sin2 b
Þ 2 cos2 q = 3 sin 2q (given sin2 b = 3 sin 2q)
\d= (2i - 2 j + 3k ) × (i + 5 j + k ) - (-5) Þ 5 cos 2q = 3
12 + 52 + 12 ,
28. (c) : From the given lines we have
d = 10 . x -1 = y+3 = z - 1 ... (A)
3 3 1 l l = s
88 JEE MAIN CHAPTERWISE EXPLORER
and x - 0 = y - 1 = z - 2 ...(B) 32. (d) : Concept using angle between the faces is equal to the
1 2 -2 angle between their normals. uuur uuur
\ Vector b to the face OAB is OA ´ OB
As lines (A) and (B) are coplanar
=uu ur5i u–u urj – 3k and vector b to the face ABC is
1 -4 -1 AB ´ AC = i –5j – 3k
\ 1 -l l = 0
\ Let q be the angle between the faces OAB and ABC
1 2 -2 (5i - j - 3k) . (i - 5 j - 3k )
Þ (2l – 2l) + 4(–2 –l) –1(2 + l) = 0
\ cos q = 5i - j - 3k i - 5 j - 3 k
Þ 5l = –10 \ l = –2
29. (b) : Using fact, two lines 19 \ q = cos -1 æè 19 ö
cos q = 35ø
x- x1 y- y1 z- z1 35
=
= and 33. (b) : The radius and centre of sphere
a1 b1 c1
x 2 + y 2 + z 2 + 2x – 2y – 4z – 19 = 0 is
x - x2 = y - y2 = z - z 2 are coplanar if 12 + 12 + 4 + 19 = 5 and centre (–1, 1, 2)
a2 b2 c2
PB ^ from centre to the plane
x2 - x1 y2 - y1 z2 - z1
a1 b1 c1 = 0 - 1 + 2 + 4 + 7
a2 b2 c2 = 4
1 -1 -1 1 + 22 + 22
Þ 1 1 - k = 0 Now (AB) 2 = AP2 – PB2
= 25 – 16
k 2 1 = 9
\ AB = 3
Þ k 2 + 3k = 0 34. (b) : In order to determine the shortest distance between
the plane and sphere, we find the distance from the centre
k = 0 or k = –3 of sphere to the planeRadius of sphere
30. (c) : Given lines can be written as –2, 1, 3
x - b = y - 0 = z - d and
a 1 c
x - b¢ = y - 0 = z - d
a¢ 1 c¢
\ Required condition of perpendicularity is
aa¢ + cc¢ + 1 = 0 r r 12x + 4y + 3z – 327 = 0
(1, b, b 2 ) ,
31. (a) : As vectors (1, r ar 2 ) , (1, r cr 2 ) are non \ Centre of sphere is (–2, 1, 3)
a, c,
coplanar. P Required distance is (2)2 + 12 + 32 + 155
1 a a 2 A B - 24 + 4 + 9 - 327
... (A) -
\ 1 b b 2 ¹ 0 122 + 42 + 32
= 26 – 13 = 13 units.
1 c c2 35. (c) : Now equation of the plane through (a, 0, 0) (0, b, 0)
(0, 0, c) is
a a2 a 3 + 1
now b b2 b 3 + 1 = 0 x y z ...(*)
c c2 c3 + 1 Þ + + = 1
xIntercept yIntercept zIntercept
On solving, we get x y z
1 a a 2 Þ + + = 1
Þ (1 + abc) 1 b b 2 = 0 abc
1 c c 2
So the distance from (0, 0, 0) to this plane to the plane (*)
is given by
d 1 = 0 + 0 + 0 - 1 1
= 1 + 1 + 1
Þ (1 + abc) = 0 by using (A) a2 b2 c2
1+1+1
a2 b2 c2
Three Dimensional Geometry 89
Similarly, d2 = 1 36. (b) : Let DR’s of normal to plane are a, b, c
1 1 1
z a¢2 + b¢2 + c¢ 2 \ a(x – 1) + b(y) + c(z) = 0 ...(*)
Þ a(0 – 1) + b(1) + c(0) = 0
(0, 0, c ) (by using (0, 1, 0) in (*))
or (0, 0, c¢ )
(0, 0, 0) Þ – a + b = 0 Þ a = b
Also angle between (*) and x + y + 0z = 3 is p/4
d = ? \ p
cos =
k 4
(a , 0, 0)
or (a¢ , 0, 0) ax +by +cz =1 a+a = 2 a
(0, b , 0) 12 + 12 a2 + b2 + c2 2 2a 2 + c2
y or (0,b¢, 0)
Now d1 = d2 given (as origin is same) Þ 2a2 + c 2 = 4a 2
Þ c = ± 2 a
1 1 \ DR’s a, b, c i.e. a, a, ± 2a
Þ = \ Required DR’s are 1, 1, 2 or 1, 1, – 2
1 1 1 1 1 1 Hence 1, 1, 2 match with choice (b)
a2 + b2 + c2 a¢2 + b¢2 + c¢2
1 1 1 1 1 1
Þ a2 + b2 + c2 - a¢2 - b¢2 - c¢ 2 = 0
90 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER VECTOR ALGEBRA
14
® ^ ^ ® ^ ^ ^ (a) - i^ + j^ - 2 k^ (b) 2 i^ - j^ + 2 k^
1. If the vectors AB = 3 i + 4 k and AC = 5 i - 2 j + 4 k are the (c) i^ - j^ - 2 k^ (d) i^ + j^ - 2 k^
sides of a triangle ABC, then the length of the median through (2010)
A is ar = i^ - j^+ 2k^, r 2i^+ 4 j^+ 4k^ and cr = l i^+ j^ + m k^
b
(a) 45 (b) 18 (c) 72 (d) 33 7. If the vectors =
(2013) are mutually orthogonal, then (l, m) =
® ^ ^ (a) (–3, 2) (b) (2, –3)
2. Let AB = 3 i + 4 k be two unit vectors. If the vectors (c) (–2, 3) (d) (3, –2) (2010)
r
r = a^ + 2 b^ and d = 5 a^ + 4 b^ are perpendicular to each other, 8. Let r j^ - k^ and cr = i^ - j^ - k^ . Then the vector satisfying
c a=
then the angle between is rrr ar × r
a´b+c =0 b
p p p p and = 3 is
3 4 6 2
(a) (b) (c) (d) (a) - i^ + j^ - 2 k^ (b) 2 i^ - j^ + 2 k^
(2012) (c) i^ - j^ - 2 k^ (d) i^ + j^ - 2 k^
uuur r uuuur r (2010)
AB q, AD = p
3. Let ABCD be a parallelogram such that = and If the vectors ar = i^ - j^+ 2k^, r 2 i^+ 4 j^+ 4k^ and cr =
ÐBAD be an acute angle. If rr is the vector that coincides b= l i^ + j^ + m k^
9.
with trrh ies aglitviteund eb yd irected from the vertex B to the side AD, are mutually orthogonal, then (l, m) =
then
r ×rqr) pr (a) (–3, 2) (b) (2, –3)
3(rp
(a) r r æ r r ör (b) r r (c) (–2, 3) (d) (3, –2) (2010)
r q èç pr × qr ø÷ p r -3q
= - p× p = + (p × p) r vr, wr are noncoplanar
u,
r ×rqr) pr 10. If vectors and p, q are real numbers,
3(rp
r r r r æ rr ör then the equality r r qur ] =
(c) r = 3q - (d) r -q èç pr × qr ÷ø p (2012) [3ur r pwr ] r w r r qv
(p × p) = + p×p pv - [ pv qu ] - [ 2w 0 holds for
4. If r = 1 æ 3 i^ + k^ ÷øö and r 1 æ 2 i^ + 3 ^j - 6 k^ ø÷ö , then the value (a) exactly two values of (p, q)
a 10 èç b= 7 çè (b) more than two but not all values of (p , q)
(c) all values of (p, q)
r br) 2br) ûù is
of ( r - b ) × éë( r ´ ´ ( r + (d) exactly one value of (p, q) (2009)
2a a a r r r
(a) 5 (b) 3 (c) –5 (d) –3 11. Tarh=e 8brn aonnd zcre =ro- 7brv.e cTthoerns a, b and c are rrelaterd b y
(2011) the angle between a and c is
5. The vectors ar and r are not perpendicular arndr cr and r are (a) p (b) 0 p p
6. b rr r r d the (c) 4 (d) 2
b ´ c = b ´ d
two vectors satisfying : and a × d = 0. Then (2008)
vector r is equal to 12. The vector ar = a i^ + 2 j^ + ^
d nd r
r c= ^ ^ b k lies in the plane of the vectors
(a) r + æ br r ö r (b) r - æ r r ö r r i^ + j^ a and bisects the angle between
b èç a × cr ø÷ c c çè ra × cr ø÷ b b= j+ k
× b a × b r
r b and cr . Then which one of the following gives possible values
r br r r r r r
(c) b - æ a × cr ö r (d) c + æ ra × cr ö b (2011) of a and b?
çè × b ÷ø c çè a × b ø÷ satisfying (a) a = 1, b = 1
(c) a = 1, b = 2
Lret rar = j^ - k^ ananddar cr× br==i^ - j^ - k^ . Then the vector r (b) a = 2, b = 2 (2008)
r 3 is b (d) a = 2, b = 1
a´b+c =0
Vector Algebra 91
13. If uˆ and vˆ are unit vectors and q is the acute angle between (a) 2 (b) 2
3 3
them, then 2uˆ ´ 3v ˆ is a unit vector for
1 (d) 2 3 2 .
(a) no value of q (c) 3 (2004)
(b) exactly one value of q r r r r r r
(c) exactly two values of q 22. Let u, v, wr be sucrh that u = 1, v r= 2, w r= 3. Irf thre
(d) more than two values of q (2007) projection v along u is equal to that of rw arlongr u and v, w
14. The values of a, for which the points A, B, C with position are perpendicular to each other then u - v + w equals
vectors 2iˆ - ˆj + kˆ, iˆ - 3 ˆj - 5k ˆ and aiˆ - 3 ˆj + kˆ respectively are
(a) 14 (b) 7 (c) 2 (d) 14.
the vertices of a rightangled triangle at c are (2004)
23. r r r are nroncorplanrar verctorsr and is a real number,
(a) 2 and 1 (b) –2 and –1 If a, b , c l r
(c) –2 and 1 (d) 2 and –1. (2006) then the vectors a + 2b + 3c, lb + 4c and (2l -1)c are non
rr r r r
15. If (a ´ b ) ´ c r (b ´r cr ) , wbrh× ecrr ¹e 0ar, , b and cr are any three coplanar for
vectors such = a´ r b ¹ 0, r and cr
that a× then a are (a) all except two values of l
(a) inclined at an angle of p/3 between them (b) all except one value of l
(b) inclined at an angle of p/6 between them (c) all values of l
(c) perpendicular (d) no value of l. (2004)
(d) parallel. (2006) 24. A particle is acted upon by constant forces 4iˆ + ˆj - 3k ˆ and
3iˆ + ˆj - kˆ which displace it from a point iˆ + 2 ˆj + 3 kˆ to
16. If r r r are noncoplanar vector and l is a real number
a, b , c r r r r
b ) l 2 b b b ùû
then éël ( r + r ûù = éëar + cr for the point 5iˆ + 4 ˆj + kˆ. The work done in standard units by
a lc
(a) no value of l the forces is given by
(a) 25 (b) 30
(b) exactly one value of l (c) 40 (d) 15.
(c) exactly two values of l r r r (2004)
(d) exactly three values of l. (2005) 25. Let a, b and c be three nonzero vre ctorsr such that no two
17. Let a, b and c be distinct nonnegative numbers. If the vectors of these are collinear. If the vector a + 2 b is collinear with
aiˆ + aˆj + ckˆ, iˆ + kˆ and ciˆ + cˆj + bkˆ lie in a plane, then c is r r r r
c and b + 3c is crollinera r with a (l being some nonzero
(a) the arithmetic mean of a and b r
(b) the geometric mean of a and b scalar) then a + 2b + 6 c equals
r r r
(c) the harmonic mean of a and b (a) l c (b) lb (c) l a (d) 0.
(d) equal to zero. (2005) (2004)
r
18. Let r iˆ - kˆ, b = xiˆ + ˆj + (1- x) kˆ and 26. r r , cr are three vectors, s uch t h a t rr cr = 0,
a= r a, b cr | = 3, then rr r r r a +b +
b , r a×b + b × c
r yiˆ + xˆj + (1+ x - y ) kˆ . Then éëar, cr ûù depends on |b | = 2, | + c× ar is equal to
c=
(a) only x (b) only y (a) –7 (b) 7 (c) 1 (d) 0.
(c) neither x nor y (d) both x and y. (2005) (2003)
19. For any vector r the value of ( r ´ iˆ)2 + r ˆj )2 + ( r kˆ ) 2 27. The vectors uuur 3iˆ + 4k ˆ and uuur 5iˆ - 2 ˆj + 4 kˆ are the sides
a , a (a ´ a´ AB = AC =
i(sa )e qaur a2 l to (b) 3ar 2 of a triangle ABC. The length of the median through A is
(c) 4ar 2 (d) 2ar 2 .
(a) 72 (b) 33 (c) 288 (d) 18.
(2005) (2003)
20. If C is the mid point of AB and P is any point outside AB, 28. Let ur = iˆ + ˆj, vr = iˆ - ˆj and wr = iˆ + 2 ˆj + 3kˆ . If nˆ is a unit
vector such that ur × nˆ = 0 and vr × nˆ = 0 , then | wr × nˆ | is equal
then uur uur uuur uur uur uuur r
(a) PuuAr + uPuBr + PuCuur= 0 (b) PuuAr + PuuBr + 2PuCuur= 0 to
(c) PA + PB = PC (d) PA + PB = 2 PC (2005) (a) 1 (b) 2 (c) 3 (d) 0.
r r r r r rwr (2003)
ur, v
21. Let a, b and c be nonzero vectors such that 29. Ifr +v arnd vra)r´e (vrt h-rwer e) noncop lana r vectors, t h e n
rr (u equals
(a ´ b) ´ r 31cr , brthecrn ars. i nIqf qe qisu athlse acute angle between the - w) × (u - vr
vectors r c= r r wr r r
and (a) u × v ´ (b) u × w ´
b
(c) 3u × r ´ wr (d) 0. (2003)
u
92 JEE MAIN CHAPTERWISE EXPLORER
30. Consider A, B, C and D with position vectors 7iˆ - 4 ˆj + 7kˆ , 34. If | r | = 5, | r | = 4, | cr | = 3 thus what will be the value of
a b
r r r
iˆ - 6 ˆj + 10kˆ , -iˆ - 3 ˆj + 4k ˆ and 5iˆ - ˆj + 5k ˆ respectively. Then r × b + b × r + r × ar , given that r + b + cr = 0
a c c a
ABCD is a (a) 25 (b) 50 (c) –25 (d) –50.
(a) rhombus
(2002)
(b) rectangle br, rcr r
r r b cr =
(c) parallelogram but not a rhombus 35. If a, are vectors show t h a t a + + r 0 and
r 5, cr = 3 then angle b and cr
(d) square. (2003) a = 7, b = between vector
r r
31. If r ´ b = b ´ r = r ´ ar then r + r + cr = is
a c c a b
(a) abc (b) –1 (c) 0 (d) 2. (a) 60º (b) 30º (c) 45º (d) 90º.
(2002) (2002)
32. ar = 3iˆ - 5 ˆj r 6iˆ + 3 ˆj are two vectors and cr is a vector 36. If r r cr are vectors such that [ar r cr ] = 4 then
r =anard ´bbr = r a, b, b
b r rr
such that c then r : : cr = [ar ´ b b´c r ´ ar ] =
a c
(a) 34 : 45 : 39 (b) 34 : 45 : 39 (a) 16 (b) 64 (c) 4 (d) 8.
(c) 34 : 39 : 45 (d) 39 : 35 : 34. r (2002)
b
r (2002) 37. If ar = 4, = 2 and the angle between ar and r is p then
b b 6
r (ar ´
33. 3lc + 2m ) = 0 then ( ar ´ r ) 2 is equal to
b
(a) 3l + 2m = 0 (b) 3l = 2m
(a) 4ar8 (b) 16
(c) l = m (d) l + m = 0. (2002) (c) (d) none of these (2002)
Answer Key
1. (d) 2. (a) 3. (d) 4. (c) 5. (b) 6. (a)
7. (a) 8. (d) 9. (a) 10. (a) 11. (a) 12. (a)
13. (b) 14. (a) 15. (d) 16. (a) 17. (b) 18. (c)
19. (d) 20. (d) 21. (d) 22. (a) 23. (a) 24. (c)
25. (d) 26. (a) 27. (b) 28. (c) 29. (a) 30. (*)
31. (c) 32. (b) 33. (b) 34. (a) 35. (a) 36. (a)
37. (b)
Vector Algebra 93
1. (d) : uuuur = 1 ( uuur + uuur Also r r Þ 2l + 4 + 4m = 0
AM AB AC) b and c are orthogonal
2 Solving the equation we get l = –3, m = 2.
r r
= 1 {(3, 0, 4) + (5, - 2, 4)} 8. (d) : We have r mb r = r b cr ] for scalars l, m, n.
2 [la nc ] lmn[a
r rr rr ar br]
= 1 (8, - 2, 8) = (4, - 1, 4) Also [a b c] = [b c r = r (cyclic)
2 a] [c
uuuur
And rrr r cr br] ( Interchange of any two vectors)
\ AM = 42 + 12 + 42 = [a b c] = -[a
33 r r r r r r r r r
[3 u pv pw] - [ pv w qu] - [2w qv qu] = 0
r r
2. (a) : c = a$ + 2b$ , d = 5a$ - 4b$ Þ 3p2 r r r - r r r + 2q2 [ur r wr ] = 0
[u v w] pq[u v w] v
\ cr ∙dr = 0 Þ (a$ + 2b$)∙ (5a$ - 4b$) = 5 - 4b$∙a$ + 10b$∙a$ - 8
Þ (3 p2 - pq + 2q2 )[ur r wr ] = 0 wr ]
r vr, wr are v r
u, r v
Þ 6 b$∙a$ - 3 = 0 Þ b$∙a $ = 1 \ q = p As noncoplanar, [u ¹ 0
23
Hence 3p2 – pq + 2q 2 = 0, p, q Î R
uuur uuuur
3. (d) : rr = BA + AQ As a quadratic in p, roots are real
r uuur uuuur Þ q 2 – 24q 2 ³ 0 Þ –23q 2 ³ 0
= - q + projection of BA across AD Þ q2 £ 0 Þ q = 0
rr r
= - r + ((ppr∙q∙pr) ) p And thus p = 0
q
r r r Thus (p, q) º (0, 0) is the only possibility.
b) b) + 2b )}
4. (c) : r - × r ´ ´ r 9. (a) : We have ar ´ r + cr = 0
(2a {(a (a b
rr rr r rr r Multairp´ly(iarn´gb rv)e+carto´ricra =lly0 with , we have
= (2a - b)× {(a ´ b ) ´ a + 2(a ´ b ) ´ b}
r r r r r r rr r rr r rr
= (2a - b )× {(a × a) b - (a × b )a + 2(a × b)b - 2(b × b )a} r rr r rr r cr =
(a b )a (a a)b a
rr r r rr rr Þ × - × + ´ 0
= (2a - b)×(b - 2a) = - 4a × a - b ×b = - 5
r cr = j^- k^ ) (i^ j^- k^ ) = -2 i^ - j^- k^
ar × r a ´ ( ´ -
b r
5. (b) : ¹ 0 (given) Thus, 3( j^- k^ ) - 2b - 2 i^- ^j - k^ = 0
ar × r = 0 r r r r \ r - i^+ j^- 2 k^
d d (b (b d) b=
r r
Now, b ´ r = b ´ Þ r ´ ´ r = ar ´ ´ 10. (a) : ar = i^- ^j + k^ , r i^+ j^+ k^ ,
c a c) b=
rr r rr rrr (ar .br)dr 2 2 4 4
(a.c)b (a.b)c (a.d)b
Þ - = - Acarr l=asnol di^br +cra aj^n+rdem cork^ ratrheo ogrotnhaolg o Þnal arÞ× cr 2=l0 +g 4iv +in g4ml =– 01 + 2m = 0
Þ rr r rr r + (ar.br)cr
Þ d(ra=.b-)d(a(r=ar× -.crbr()) abr.c+)cbr
r r cr = Solving the equation we get l = –3, m = 2.
a b
6. (a) : We have ´ + 0 11. (a) : ar = r
8 b
Multarip´ly(iarn´gb rv)e+carto´ricra =lly0 with ar , we have carr =an-d7 b brr
rr rr are parallel and r and cr are antiparallel.
r b )a r a)b r cr = b
Þ (a × - (a × + a ´ 0 HTheunsc ea rthaen da ncrg lea rbee tawnteiepna raarllaenl.d cr is p.
r ´3(cr j^=- (k^j^)--k^2)br´ -( i^2-i^ j^- k^ ) = -2 i^ - j^- k^
a - ^j - k^ = 0
Thus, r r cr . ar
a b
r i^+ j^- k^ 12. (a) : r lies in the plane of and Also bisects the
b= b r
\ - 2 r angle and r Thus r = l(b + r
ar = b= c . a c )
7. (a) : i^- ^j + 2 k^ , 2 i^+ 4 ^j + 4 k^ , cr = l i^+ j^+ m k^
æ^ ö^ ^ æ ^ ö^ ^
r r Þ r × cr = 0 giving ^ ^ ^ lçi + ^ + j + k ÷ = l ç i + 2 j + k ÷
a and c are orthogonal a
a i + 2 j + b k = j
l – 1 + 2m = 0 è2 2 ø è 2 ø
94 JEE MAIN CHAPTERWISE EXPLORER
on comparison, l = 2a, l = 2 and l = 2 b 19. (d) : Let ar = a1i + b1 j + c1 k r r
Thus a = 1 and b = 1 ar 2r = a12 + c12 a´i
\ (a ´ ra12 + b12 + c12 \ = - b1k + c1 j
\ i ) 2
ar= ´b12ˆj )+ 2 c12
13. (b) : | 2uˆ ´ 3vˆ |= 1 Þ 6 | uˆ || vˆ || sin q |= 1 Þ sin q = 1 Similarly ( =
6
(ar ´ kˆ ) 2 = a12 + b12
2uˆ ´ 3v ˆ is a unit vector for exactly one value of q.
uuur uuur
14. (a) : Now CA × CB = 0 r iˆ)2 r ˆj )2 (ar ´ kˆ )2 2( a12 b12 c12 )
\ ( a ´ + (a ´ + = + +
= 2ar 2 .
A
(2 iˆ - ˆj + kˆ )
20. (d) : Let P is origin P(0)
Let uur r uur r
PA = a, PB = b
\ uuur rr
PC = a + b
( aiˆ - 3 ˆj + kˆ ) 2 ® ® ® ®
A(a) a + b B(b)
uur uur ar r C
PA + PB = b
B ( iˆ - 3 ˆj + kˆ ) C Now + 2
uur r r uuur
whereu uCr A = (2 - a)iˆ - 2 ˆj a +2 b øö÷ 2 PC .
and CB = (1 - a)iˆ - 6kˆ = æ =
2èç
Þ a 2 – 3a + 2 = 0 Þ (a – 2)(a – 1) = 0 rr r 1 b c ar (As given)
a´b c
Þ a = 1, 2 r r ( ) 21. (d) : ´ =
´b ´(b ´
Grivern r(ar r r r r 3 r
15. (d) : )r´ c =a Þ c ) r rr b
Þ r r r r b b ×c ar 1 r r
(b ×c )ar = (a × b ) c r = r ( ) Þ ( a × c ) - = 3 c a
r r l1a l 2 c
r
Þ (arl 1|| =crb × c ,l2 = a × b are scalar quantities) Þ r × cr = 1 b r Þ cos q = –1/3
-b c
3
16. (a) : From given sin q = 8 2 2
=
r rr r r r r 93
2 ( r b (lb ´ lc ) a × (b + c ) ´ b rr r wr
l a + ) × = v ×u u × r wr =
v
l 2 ar r r r r r r (cr ´ r 22. (a): Given = and × 0
(lb lc ) l2b (lb c) a b ) uu
Þ × + + × ´ = ×
Þ l4 [a b c] = - [a b c] Þ l4 + 1 = 0 Also ur = 1, vr = 2, wr = 3
Þ (l2 )2 + 1 = 0 D < 0 Now r - r + wr 2
u v
Þ No value of l exist on real axis. = r 2 + r 2 + wr 2 - r × r - r × r + r × wr
u v 2u v 2v w 2u
17. (b) : We are given that points lies in the same plane. We know
that the vector L, M, N are coplanar if = 1 + 4 + 9 + 0 = 14
23. (a) : Using the condition of coplanarity of three vectors
r r a a c 12 3
L × (M ´ N ) = 0 Þ 1 0 1 = 0 Þ c = ab \ 0 l 4 = 0
ccb 0 0 2l - 1
\ C is G.M. of a and b.
1
18. (c) : [a, b, c] = r r cr ) = a1 a2 a3 Þ l = 0, .
a × (b ´ b1 b2 b3
2
c1 c2 c3 r r r
1 0 -1 24. (cr) : Trotal fro rce F = F1 + F2 = 7i + 2j – 4k and displacement
d = d2 - d1
= x 1 1 - x C 3 ® C3 + C1
y x 1 + x - y = (5 – 1)i r+ (4r – 2)j + (k – 3k)
\ W.D = F × d = 28 + 4 + 8 = 40
\(adn )d:a rb rA ++s 23ab rcr r +=is 2Pcb rocrl liinse caor lwliniteha ra rw \it hbr
25. r
c
1 0 0 = Qar ...(i)
= x 1 1 = 1(1) = 1 r
+ 3c
y x 1 + x
which is independent of x and y. Naro-w6 b cry =(i ) Pacnrd- (2ii Q) arw e have
Vector Algebra 95
Þ ar (1 + 2Q) + cr (–6 – P) = 0 ( ) 31. (\ÞÞSci)m a:ri l ´aIfrba ablrrrrpy ´+o ×=sbbrcrrs ´ibbr=b=rclr´ e + ccr-rs=c raa´r y =cra r =´arcr a+r´-bararr +´cr = 0
r
Þ 1 + 2Q = 0 and – P – 6 = 0 a
Q = –1/2, P = –6
Paru+tti2nbgr +th6esc re = v a0l ue either in (1) or in (ii) we get
r
26. (a) : r +b + cr = 0
a
Consider (a + b +2 (car)2 . br + r .cr + r . ar) rrr
= a 2 + b 2 + c 2 + b c i j k
cr rr 3 -5 0
r r r .cr r . ar ) - (a 2 + b2 + c 2 ) 32. (b) : Given = a´b =
Þ (a .b b c 2
+ + = cr r 6 3 0
k
(12 + 22 + 32 ) \ = 39 r
= - 34, b =
Now ar = 45 and
2 rr
= –7 c = 39k = 39
27. (b) : Median through any vertex divide the opposite side into \ r : r : cr =
a b
tuwuuor equuuaurl parts uuur 34 : 45 : 39
r
AB + AC = 2AD 33. (b) : 3l cr = 2m (b ´ r r r ar
a) c || b
Þ uuur = 1 [ uAuBur + uuur Þ either 3l = 2m or ´
AD 2 AC ]
but 3l = 2m
1 :|22 arr5(War|+2er×1b +r6h|+a+brrvbr9e|2r× +cr+ar2+|+r(cracrrbr|r×2× + barr+ )c+r2==b(r-ar×0r 5c×rb0rÞ++crbr(× aarr× c r)+=+br 0cr+× cr )2 = 0
= [8i – 2j + 8k] r
2 uuur 34. (a) a ) = 0
Þ
\ AD = 33 Þ
Þ
28. (c) : nˆ rr r³ ur . nˆ = 0 = r . nr
|| u ´r v v
now nˆ = ur´ vrv | Þ (a ×rb +rb ×c +c × a ) = - 25
u r b +b× rr ar ) = 25
c+c×
1´ 1 (- 2kˆ ) - kˆ \ (a ×
= 2 2 = (bra )a n: dGcri vseon c oarns+idbrer+
crr
Now wr . nˆ = (i + 2 ˆj + 3kˆ) . (- kˆ ) 35. b = r0, wer need angle between
+c = -a
= |–3| = 3 Þ b2 + c2 + 2|b| |c| cos q = a2
r r r . [ur r r r r wr ] a2 - b2 - c 2 49 - 25 - 9 1
29. (a)Q : vr(u´ +r v - w ) ´ v - u ´ w + v ´
v = Þ cos q = ==
0 2b c 2 ´ 5 ´ 3 2
r .(ur ´ r - r .(ur ´ r + r . (vr ´ r v \ q = 60° r rr r r arr ]
u v) u w) u w ) u c b´
-wr+.(vrur.u´r ´vr )vr+-wrvr..((urur´´wrwr))-+wrv .. ((vvrr r k
´´ wwr )) 36. =(=a )(a ra:r ´´Cbrobr.n)[s(.[ ikrdkr.ea´rr )(cr[cra-´´(arkr b).] c rbw)arh´ ]e rce ´ cr
=
= r . r ´ r + r . ( r ´ r - r . (ur ´ r (ëé(aaarrr ´.´(bbbrrr))´ ..ëécr [[)((ùûbbréë´´crc.cr ())ar..a a´r]]crbcrr )-- ûù[0(=b r =´ëé carr)(. b(.r cbr´]´a rcr ùûcr) ). aûùr2 =ëé ( a1r6´
u (v w) v w u) w v )
r r r r r r r . (vr r
= u . (v ´ w ) + u . (v ´ w) - u ´ w ) ( ) =
r r wr )
= u . (v ´ (³ [a b c] = [b c a] = [c a b]) = br) . cr ùû
=
uuur uuur uuur
30. (*) : AB = OB - OA = 6iˆ + 2 ˆj - 3k ˆ
uuur
\ | AB |= 49u uur= 7
Similarly BC = 2iˆ - 3 ˆj + 6 kˆ 37. (b) : Using fact: (ar ´ r = a 2b 2 – ( ar r
b ) 2 . b ) 2
uuur = a 2b 2 – a2 b 2 cos 2q
\ =| C uBu-DuCr2 i ˆ|= + =3- ˆj6 -iˆ4 -29 k2ˆ = ˆj| -uD7u Akurˆ | =| Cu uDur1 |7 = p
41
uuur = (4 × 2) 2 – (4 × 2) 2 cos 2
DA 6
p 1
= 64 × sin2 = 64 × = 16
6 4
96 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER STATISTICS
15
1. All the students of a class performed poorly in Mathematics. 6. If the mean deviation of number 1, 1 + d, 1 + 2d, ...., 1 + 100d
The teacher decided to give grace marks of 10 to each of the from their mean is 255, then the d is equal to
students. Which of the following statistical measures will not (a) 20.0 (b) 10.1
change even after the grace marks were given? (c) 20.2 (d) 10.0 (2009)
(a) median (b) mode 7. The mean of the numbers a, b, 8, 5, 10 is 6 and the variance
(c) variance (d) mean (2013)
is 6.80. Then which one of the following gives possible values
2. Let x 1 , x 2 , ...., x n be n observations, and let be their arithmetic of a and b?
mean and s2 be their variance. (a) a = 3, b = 4 (b) a = 0, b = 7
Statement 1 : Variance of 2x 1 , 2x 2 , ...., 2x n is 4s2 . (c) a = 5, b = 2 (d) a = 1, b = 6 (2008)
Statement 2 : Arithmetic mean of 2x 1, 2x 2 , ...., 2x n is 4.
(a) Statement 1 is true, Statement 2 is true; Statement 2 is 8. The average marks of boys in class is 52 and that of girls
not a correct explanation for Statement 1. is 42. The average marks of boys and girls combined is 50.
(b) Statement 1 is true, Statement 2 is false. The percentage of boys in the class is
(c) Statement 1 is false, Statement 2 is true. (a) 80 (b) 60
(d) Statement 1 is true, Statement 2 is true; Statement 2 is (c) 40 (d) 20. (2007)
a correct explanation for Statement 1. (2012)
9. Suppose a population A has 100 observations 101, 102, ...,
3. If the mean deviation about the median of the numbers a, 2a,
200, and another population B has 100 observations 151, 152,
......., 50a is 50, then |a| equals
..., 250. If V A and V B represent the variances of the two
(a) 4 (b) 5 (c) 2 (d) 3
(2011) populations, respectively, then V A /V B is
4. For two data sets, each of size 5, the variances are given to (a) 1 (b) 9/4 (2006)
be 4 and 5 and the corresponding means are given to be 2 (c) 4/9 (d) 2/3.
and 4, respectively. The variance of the combined data set is 10. Let x 1 , x 2 , ...., x n be n observations such that
å x12 = 400 and å xi = 80. Then a possible value of n among
(a) 5 (b) 11 (c) 6 (d) 13 the following is
2 2 2
(2010) (a) 18 (b) 15
5. Statement1 : The variance of first n even natural numbers (c) 12 (d) 9. (2005)
is n2 - 1 11. In a frequency distribution, the mean and median are 21 and
4
22 respectively, then its mode is approximately
Statement2 : The sum of first n natural numbers is
n(n + 1) and the sum of squares of first n natural numbers is (a) 20.5 (b) 22.0 (2005)
2 (c) 24.0 (d) 25.5.
n(n + 1)(2n + 1) 12. In a series of 2n observations, half of them equal a and
6 remaining half equal – a. If the standard deviation of the
observations is 2, then |a| equals
(a) Statement 1 is true, Statement 2 is true; Statement 2 is
not a correct explanation for Statement 1.
(b) Statement 1 is true, Statement 2 is false. (a) 2 (b) 2
(c) Statement 1 is false, Statement 2 is true.
(d) Statement 1 is true, Statement 2 is true; Statement 2 is 1 (d) n2 .
n
a correct explanation for Statement 1. (2009) (c) (2004)
Statistics 97
13. Consider the following statements : 15. In an experiment with 15 observations on x, the following
results were available.
(1) Mode can be computed from histogram
(2) Median is not independent of change of scale å x2 = 2830, å x = 170
(3) Variance is independent of change of origin and scale.
Which of these is/are correct? One observation that was 20 was found to be wrong and was
(a) only (1) and (2) (b) only (2) replaced by the correct value 30. Then the corrected variance
(c) only (1) (d) (1), (2) and (3). (2004) is
14. The median of a set of 9 distinct observations is 20.5. If each (a) 188.66 (b) 177.33
of the largest 4 observations of the set is increased by 2, then (c) 8.33 (d) 78.00. (2003)
the median of the new set 16. In a class of 100 students there are 70 boys whose average
(a) is decreased by 2 marks in a subject are 75. If the average marks of the complete
(b) is two times the original median class is 72, then what is the average of the girls?
(c) remains the same as that of the original set (a) 73 (b) 65
(d) is increased by 2. (2003) (c) 68 (d) 74. (2002)
Answer Key
1. (c) 2. (b) 3. (a) 4. (b) 5. (c) 6. (b)
7. (a) 8. (a) 9. (a) 10. (a) 11. (c) 12. (a)
13. (a) 14. (b) 15. (d) 16. (b)
98 JEE MAIN CHAPTERWISE EXPLORER
1. (c) : 1 st solution : Variance doesn’t change with the change = 1 (40 + 105) - 9 = 145 - 90 = 55 = 11 .
10 10 10 2
of origin. 1
n 2n d solution :
s12 = (xi - x )2 s12 = 4, n1 = 5, x1 = 2
å2 nd solution : s22 = 5, n2 = 5, x2 = 4
1 (xi + 10) - (x + 10) 2 Hence s 12 = s2 2
å { } s22 =n
2. (b) : x 1, x2 , x 3 , .... x n , A.M. = x , Variance = s 2
Statement 2 : A.M. of 2x 1, 2x 2, ...., 2x n x12 = n1x1 + n2x2 = 5 ´ 2+5´ 4 = 3
n1 + n2 10
= 2(x1 + x2 + ..... + xn ) = 2x d1 = (x1 - x12 ) = -1, d2 = (x2 - x12 ) = 1
n
Given A.M. = 4x \ Statement 2 is false. s = n1s12 + n2s22 + n1d12 + n2d22
n1 + n2
3. (a) : Median is the mean of 25 th and 26 th observation.
M = 25a + 26 a = 25.5 a = 5.4 + 5.5 + 5.1 + 5.1 = 55 = 11
2 10 10 2
MD(M) = å |ri - M| \ s2 = 11
2
N
Þ 50 = 1 {2| a|´ (0.5 + 1.5 + ... + 24.5)} 5. (c) : Sum of first n even natural numbers
50 = 2 + 4 + 6 + .... + 2n = 2 (1 + 2 + ... + n)
Þ 2500 = 2|a| × 225 × 25. \ |a| = 4 n(n + 1)
= 2 × 2 = n(n + 1)
4. (b) : 1 st solution: n(n + 1)
Mean (x) = n = n + 1
s12 = 4 üï x = 2
= 5ýïþ y = 4
s 2 Variance = 1 2 - ( x ) 2
2 n (å xi )
We have å xi = 2 Þ å xi = 10 = n1 (22 + 42 + .... + (2n)2 ) - (n + 1) 2
5
Similarly, å yi = 20 = n1 × 22 (12 + 22 + .... + n2 ) - (n + 1) 2
æ 1 xi2 ö 2 4 = 1 xi2 - 4
çè 5 ø÷ 5
å å s12 = - x Þ = 4 × n(n + 1)(2n + 1) (n + 1) 2
n 6 -
Þ å1 xi2 = 8. \ å xi2 = 40. = 32 × (n + 1)(2n + 1) - (n + 1) 2
5
2 æ 1 yi2 ö 2 5 = 1 yi2 - 16 (n + 1)
2 èç 5 ø÷ 5 = 3 [2(2n + 1) - 3(n + 1)]
å å s = - y Þ
1 = (n + 1) × (n - 1) = n 2 - 1
5 3 3
yi2 = 21. \ yi2 = 105
å åÞ
( ) å å s2 = 1 xi2 + yi2 - æx + y ö2 6. (b) : The numbers are 1, 1 + d, 1 + 2d, ..., 1 + 100d.
10 èç 2 ø÷ The numbers are in A.P.
Then mean = 51 st term = 1 + 50d = x (say)
Statistics 99
Mean deviation (M.D.) = 1 101 xi - x | Þ n = 17 but not given in choice.
å | \ n = 18 is correct number.
n 11. (c) : Using fact, mode = 3 median – 2 mean
i =1 = 3 × 22 – 2 × 21 = 3(22 – 14) = 3 × 8 = 24.
1
= 101 [50d + 49d + 48d + .... + d + 0 + d + 2d + ... + 50d ]
+ d + 0 + d + 2d + .... + 50d] 12. (a) : According to problem
1 X Value of X d = value of X– X (X - X ) 2
= 101 × 2d (1 + 2 + .... + 50)
x 1 a a a2
1 50 × 51 50 × 51 x 2 a a
= 101 × 2 d × 2 = 101 d – – – a 2
But M.D. = 255 (given) – – –
– – –
50 × 51 – – – –
Þ 101 d = 255 a a –
– –a –a
Þ d = 101 ´ 255 = 101 ´ 255 = 10.1 x n –a –a –
50 ´ 51 2550 xn + 1 – – –
xn + 2 – – a 2
7. (a) : The mean of a, b, 8, 5, 10 is 6 – – – a 2
– –a –a
a + b + 8 + 5 + 10 = 6 –
5 – –
xn + n
–
a 2
Þ å (X - X ) 2 = 2na2
Þ a + b + 23 = 30 Þ a + b = 7 ...(1) S X = 0
Again variance = å( xi - A ) 2 = 6.8 \ X = SX = 0 = 0
n N 2 n
(a - 6)2 + (b - 6)2 + 4 + 1 + 16 ( X - X ) 2 X 2 - ( X ) 2
= 6.8 =
Þ 5 å å Now SD =
Þ a 2 + b 2 – 12(a + b) + 36 + 21+72=5×6.8 = 34 NN
Þ a2 + b 2 – 12 × 7 + 72 + 21 = 34 2 = 2 na 2 - 0 ; 2 = a 2 ; 2 = |a|
2 n
\ a2 + b2 = 25 ...(2)
using (1) we have 13. (a) : Mode can be computed by histogram
Median will be changed if data’s are changed so (2) is correct.
a 2 + (7 – a)2 = 25 Þ a 2 + 49 – 14a + a 2 = 25 Variance depends on change of scale so (3) is not correct.
Þ a2 – 7a + 12 = 0 \ a = 3, 4 also b = 3, 4
8. (a) : Let x and y are number of boys and girls in a class 14. (b) : Total number of observations are 9 which is odd which
means median is 5 th item now we are increasing 2 in the last
respectively. four items which does not effect its value. The new median
remain unchanged.
52 x + 42 y = 50
x + y
Þ x = 4 y Þ x = 4 and x x = 4 15. (d) : åx = 170 and åx2 = 2830
y 1 +y 5
Required percentage = x x ´100 = 45 ´100 = 80%. Increase in åx = 10 and åx¢ = 170 + 10 = 180
+ y Increase in åx 2 = 900 – 400 = 500 then
åx¢ 2 = 2830 + 500 = 3330
9. (a) : Series A = 101, 102, 103, ......., 200 = 1 ´ 3330 - æ 1 ´ 180öø 2 = 222 – (12) 2 = 78
Series B = 151, 152, 153, ........., 250 15 è 15
Series B is obtained by adding a fixed quantity to each item
of series A, we know that variance is independent of change 16. (b) : Using x = ( x1 + x2 + ... + x1 00 ) = 72
of origin both series have the same variance so ratio of their 100
variances is 1.
\ x 1 + ... + x1 00 = 7200 ...(i)
10. (a) : Using well known fact that root mean square of number Again x1 + x2 + ... + x70 = 75 ...(ii)
³ A.M. of the numbers
70
x 1 + ... + x7 0 = 75 × 70
Þ 400 ³ 80 Þ 20 ³ 80 \ Average of 30 girls = 7200 - 5250 = 65
n n n n
Þ n ³ 4 Þ n ³ 16 30
100 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER PROBABILITY
16
1. A multiple choice examination has 5 questions. Each question (a) Statement1 is true, Statement2 is true; Statement2 is a
has three alternative answers of which exactly one is correct. correct explanation of Statement1.
The probability that a student will get 4 or more correct answers (b) Statement1 is true, Statement2 is true, Statement2 is
just by gues sing is not a correct explanation for Statement1.
10 17 13 11 (c) Statement1 is true, Statement2 is false. (2010)
(a) 3 5 (b) 35 (c) 3 5 (d) 35 (2013) (d) Statement1 is false, Statement2 is true.
( ) 2. 1
Three numbers are chosen at random without replacement 7. In a binomial distribution B n, p = 4 , if the probability of
from {1, 2, 3, ....., 8}. The probability that their minimum is
3, given that their maximum is 6, is at least one success is greater than or equal to 190 , then n is
(a) 1 (b) 2 (c) 3 (d) 1 greater than 9
4 5 8 5 1 (b) log10 4 - log10 3
(2012) (a) log10 4 + log10 3
3. If C and D are two events such that C Ì D and P(D) ¹ 0, 4 1
(c) log10 4 - log10 3 (d) log10 4 - log10 3 (2009)
then the correct statement among the following is
8. One ticket is selected at random from 50 tickets numbered
(a) P(C|D) < P(C) (b) P(C|D) = P(D) 00, 01, 02, …., 49. Then the probability that the sum of the
P(C) digits on the selected ticket is 8, given that the product of
(c) P(C|D) = P(C) (d) P(C|D) ³ P(C) (2011) these digits is zero, equals
4. Consider 5 independent Bernoulli’s trials each with probability (a) 1 (b) 5 (c) 1 (d) 1
of success p. If the probability of at least one failure is greater 7 14 50 14
than or equal to 31 , then p lies in the interval (2009)
32 9. A die is thrown. Let A be the event that the number obtained
(a) éêë0, 12 ûúù (b) æ 11 , 1ûúù is greater than 3. Let B be the event that the number obtained
èç 12 is less than 5. Then P(A È B) is
(c) æ 1 , 3 ù (d) æ 3 , 11 ù (2011) 2 3 (c) 0 (d) 1 (2008)
çè 2 4 úû èç 4 12 úû (a) 5 (b) 5
5. An urn contains nine balls of which three are red, four are 10. It is given that the events A and B are such that
blue and two are green. Three balls are drawn at random without P( A) = 1 , P( A | B) = 1 and P( B | A) = 32 . Then P(B) is
replacement from the urn. The probability that the three balls 42
have different colours is
(a) 1/3 (b) 2/7 (c) 1/21 (d) 2/23 1 1 1 2
(2010) (a) 2 (b) 6 (c) 3 (d) 3 (2008)
6. Four numbers are chosen at random (without replacement) 11. A pair of fair dice is thrown independently three times. The
from the set {1, 2, 3, ..., 20}. probability of getting a score of exactly 9 twice is
Statement1 : The probability that the chosen numbers when (a) 8/729 (b) 8/243
arranged in some order will form an A.P. is 815 . (c) 1/729 (d) 8/9. (2007)
Statement2 : If the four chosen numbers form an A.P., then
12. Two aeroplanes I and II bomb a target in succession. The
the set of all possible values of common difference is {±1, probabilities of I and II scoring a hit correctly are 0.3 and
0.2, respectively. The second plane will bomb only if the
±2, ±3, ±4, ±5}.
Probability 101
first misses the target. The probability that the target is hit (a) 0.35 (b) 0.77
(c) 0.87 (d) 0.50.
by the second plane is (2004)
(a) 0.2 (b) 0.7 19. The probability that A speaks truth is 4/5, while this probability
(c) 0.06 (d) 0.14. (2007) for B is 3/4. The probability that they contradict each other
13. At a telephone enquiry system the number of phone calls when asked to speak on a fact is
regarding relevant enquiry follow Poisson distribution with a (a) 7/20 (b) 1/5
average of 5 phone calls during 10minute time intervals. The (c) 3/20 (d) 4/5. (2004)
probability that there is at the most one phone call during a 20. The mean and variance of a random variable X having a
10minute time period is binomial distribution are 4 and 2 respectively, then
6 5 6 6 P (X = 1) is
(a) 5e (b) (c) (d) e 5 .
(a) 1/16 (b) 1/8 (c) 1/4 (d) 1/32.
6 55 (2006)
(2003)
14. A random variable X has Poisson distribution with mean 2. 21. Five horses are in a race. Mr. A selects two of the horses at
The P (X > 1.5) equals random and bets on them. The probability that Mr. A selected
(a) 0 (b) 2/e2 the winning horse is
(c) 3/e 2 (d) 1 - e32 . (a) 3/5 (b) 1/5
(2005) (c) 2/5 (d) 4/5. (2003)
15. Three houses are available in a locality. Three persons apply 22. Events A, B, C are mutually exclusive events such that
for the houses. Each applies for one house without consulting P( A) = 3x3 + 1, P(B) = 1 -4 x and P(C) = 1 -22 x . Then set of
possible values of x are in the interval
others. The probability that all the three apply for the same
house is
(a) 1/9 (b) 2/9 (a) êéë13, 23úûù (b) ëêé13, 133ùûú
(c) 7/9 (d) 8/9. (2005)
16. Let A and B be two events such that P (A È B) = 1 , (d) ëêé13, 12ûúù .
6
P( A) = 1 , (c) [0, 1] (2003)
P( A Ç B) = 1 a n d 4 w h er e stands for
4 A
complement of event A. Then events A and B are 23. A die is tossed 5 times. Getting an odd number is considered
(a) equally likely but not independent a success. Then the variance of distribution of success is
(b) equally likely and mutually exclusive (a) 8/3 (b) 3/8
(c) mutually exclusive and independent (c) 4/5 (d) 5/4. (2002)
(d) independent but not equally likely. (2005) 24. A and B are events such that P (A È B) = 3/4,
17. The mean and the variance of a binomial distribution are 4 and P (A Ç B) = 1/4, P (A ) = 2/3 then P (A Ç B ) is
2 respectively. Then the probability of 2 successes is: (a) 5/12 (b) 3/8
(a) 128/256 (b) 219/256 (c) 5/8 (d) 1/4. (2002)
(c) 37/256 (d) 28/256. (2004) 25. A problem in mathematics is given to three students A, B, C
18. A random variable X has the probability distribution: and their respective probability of solving the problem is
X : 1 2 3 4 5 6 7 8 1/2, 1/3 and 1/4. Probability that the problem is solved is
P(X) : 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05
For the events E = {X is a prime number} and F = {X < 4}, (a) 3/4 (b) 1/2
the probability P(E È F) is:
(c) 2/3 (d) 1/3. (2002)
Answer Key
1. (d) 2. (d) 3. (d) 4. (a) 5. (b) 6. (c)
7. (d) 8. (d) 9. (d) 10. (c) 11. (b) 12. (d)
13. (d) 14. (d) 15. (a) 16. (d) 17. (d) 18. (b)
19. (a) 20. (d) 21. (c) 22. (d) 23. (d) 24. (a)
25. (a)
102 JEE MAIN CHAPTERWISE EXPLORER
1. (d) : P(correct answer) = 1/3 The total number of ways n(S) = 20C 4
The required probability = 5C4 æ 1 ö4 æ 2 ö + 5 C5 èçæ 1 ö5 The desired probability = 57 = 57 ´ 24 = 1
çè 3 ø÷ èç 3 ÷ø 3ø÷ 20 C 4 20 ´ 19 ´ 18 ´ 17 85
Now statement2 is false and statement1 is true.
= 5´2 + 1 = 11 7. (d) : Probability of at least one success
35 35 35 = 1 – No success = 1 – nC n q n
where q = 1 – p = 3/4
2. (d) :3 numbers are chosen from {1, 2, 3, ....., 8} without
replacement. Let A be the event that the maximum of chosen we want 1 - çèæ 3 ö 4 ³ 9
numbers is 6. 4 ø÷ 10
Let B be the event that the minimum of chosen numbers is 3.
1∙1∙2 Þ 1 ³ æçè 3 ö4 Þ æèç 3 ö4 £ 1
10 4 ÷ø 4 ø÷ 10
P(B / A) = P(A Ç B) = 8 C3 = 2 = 1
P(B) 5 C2 10 5
Taking logarithm on base 10 we have
8 C3
n log10(3/4) £ log1010 –1
3. (d) : P(C|D) = P(C Ç D) as C Ì D, P(C) Ì P(D). Þ n(log103 – log104) £ –1
P(D) Þ n(log104 – log103) ³ 1
\ P(C Ç D) = P(C) Þ n³ 1
We have, P(C|D) = P(C) log10 4 - log10 3
P(D)
8. (d) : Any number in the set
As 0 < P(D) £ 1 we have P(C|D) ³ P(C) S = {00, 01, 02, ...., 49} is of the form ab where
4. (a) : Probability of at least one failure a Î {0, 1, 2, 3, 4} and b Î {0, 1, 2, ..., 9} for the product of
digits to be zero, the number must be of the form either x0
= 1 – P(no failure) = 1 – p 5 which are 5 in numbers, because x Î{0, 1, 2, 3, 4}
or of the form 0x which are 10 in numbers because
Now 1 - p5 ³ 31 x Î {0, 1, 2, ..., 9}
32 The only number common to both = 00
Thus the number of numbers in S, the product of whose digits
Þ p5 £ 1 thus p £ 1 \ p Î [0,1 / 2] is zero = 10 + 5 – 1 = 14
32 2 Of these the number whose sum of digits is 8 is just one, i.e.
08
5. (b) : n(S) = 9C 3 = 9 ´ 8 ´ 7 = 84 The required probability = 1/14.
6
n(E) = 3C1 ∙ 4C 1 ∙ 2C 1 = 3 × 4 × 2 = 24.
The desired probability = 24 = 2 . 9. (d) : A = {4, 5, 6}
84 7 Also B = {1, 2, 3, 4}
We have A È B = {1, 2, 3, 4, 5, 6} = S
6. (c) : Number of A.P.’s with common difference 1 = 17 Where S is the sample space of the experiment of throwing
Number of A.P.’s with common difference 2 = 14 a die. P(S) = 1, for it is a sure event.
Number of A.P.’s with common difference 3 = 11 Hence P(A È B) = 1
Number of A.P.’s with common difference 4 = 8
Number of A.P.’s with common difference 5 = 5 10. (c) : From the definition of independence of events
2 P( A/ B ) = P( A Ç B )
Number of A.P.’s with common difference 6 = 57 P( B)
Probability 103
Then P(B) ∙ P(A/B) = P(A Ç B) ...(1) 17. (d) : Given np = 4 and npq = 2
Interchanging the role of A and B in (1) ...(2)
P(A)P(B/A) = P(B Ç A) q = npq = 2 = 1 so p = 1 – 1/2 = 1
As A Ç B = B Ç A, we have from (1) and (2) np 4 2 2
P(A)P(B/A) = P(B)P(A/B)
Now npq = 2 \ n = 8
\ BD is given by P(X = r) = 8C r pr qn – r
1 2 1 1 2 1 \ P(X = r = 2) = 8C 2 çæè 1 ö8 28
Þ 4 × 3 = P( B) × 2 Þ P( B) = 4 × 3 × 2 = 3 2 ÷ø =
256
11. (b) : Possibility of getting 9 are (5, 4), (4, 5), (6, 3), (3, 6) 18. (b) : From the given table prime numbers are
2, 3, 5, 7
Probability of getting score 9 in a single throw = p = 4 = 91 ‘E’ denote prime
36
Required probability = probability of getting score 9 exactly ‘F’ denote the number < 4
\ P(E) = P(2 or 3 or 5 or 7)
twice =3 C2 èçæ 1 ÷øö2 ´ çæè 8 ÷öø = 284 3 .
9 9 (Events 2, 3, 5, 7 are M.E)
= P(2) + P(3) + P(5) + P(7) = .62
12. (d) : P (I) = 0.3, P ( I ) = 1 - 0.3 = 0.7, P(F) = P(1 or 2 or 3) (events 1, 2, 3 are m.E.)
P (II) = 0.2, P (II ) = 1 - 0.2 = 0.8 = P(1) + P(2) + P(3) = .50
Required probability = P ( I Ç II) = P ( I ) P(II)
P(E Ç F) = P(2 or 3) = P(2) + P(3) = .35
= (0.7)(0.2) = 0.14.
P(E È F) = P(E) + P(F) – P(E Ç F)
= .62 + .50 – .35 = .77
13. (d) : We know that poission distribution is given by 4 1
19. (a) : P(A) = 5 \ P( A ) = 5
P( x = r ) = e - l l r where l = 5
r ! 3 1
P(B) = 4 \ P (B ) = 4
Now P(x = r £ 1) = P(x = 0) + P(x = 1)
e - l l - l 6 Now we needed P(A) P (B ) + P(B) P ( A )
0! e e5
= + = e -5 (1 + 5) = . 4 1 3 1 7
1! = 5 ´ 4 + 4 ´ 5 = 20
14. (d) : P( X = r ) = e -llr (l = mean ) 20. (d) : Given mean np = 4, npq = 2
r!
npq 2 1
\ P(X = r > 1.5) = P(2) + P(3) + ... ¥ Þ np = 4 \ q = p = 2 and n = 8
= 1 – [P(0) + P(1)] 1 ö8
2 ÷ø
é e -2 e -2 ´ 22 ù 3 . Now P(X = r) = 8 Cr çèæ
ëê 2 úû e2
= 1 - + = 1- (Use p(X = r) = nC r p r qn –r
15. (a) : No. of houses = 3 = No. of favourable cases \ p(X = 1) = 8C 1 çæè 1 ö8 8 = 1
2 ÷ø = 16 ´ 16 32
No. of applicants = 3,
\ Total number of events = 33 21. (c) : No. of horses = 5
(because each candidate can apply by 3 ways) 4 3
´
Required probability = 3 = 91 . \ Probability that A can’t win the race = 54
33
Probability that ‘A’ must win the race = 1 - P( A) P(B )
16. (d) : P( A È B) = 1- P( A È B)
= 1 - P( A) - P(B) + P( A Ç B) = 1 - 12 = 2
20 5
1 = 1 - 3 - P(B) + 1
64 4 22. (d) : A, B, C are mutually exclusive
\ 0 £ P(A) + P(B) + P(C) £ 1
P(B) = 1 - 1 Þ P(B) = 4 = 1 0 £ P(A), P(B), P(C) £ 1 ...(i)
2 6 12 3 Now on solving (i) and (ii) we get ...(ii)
1 1
now P(AÇ B) = 1 = 1 ´ 3 = P( A)P(B) 3 £ x £ 2
4 3 4
so even are independent but not equally likely as
P(A) ¹ P(B).
104 JEE MAIN CHAPTERWISE EXPLORER
æ 1 ön = 240 - 25
23. (d) : n = 5 p = q = 1/2 P(X = r) = 5 C r çè 2 ÷ø 32 4
x i fi f i x i f i x i 2 40 5
= 32 = 4
(0) 0 æ 1 ö5 0 0 A B
çè 2 ÷ø 24. (a) : Given P (A È B) = 3/4
P(A Ç B) = 1/4
5C 1 çèæ 1 ö5 1 ´ 5 5
(1) 2 ÷ø 32 32 P ( A ) = 2/3
2 2 10
(2) 5C 1 çæè 1 ö5 2 ´ 10 2 P ( A Ç B) = P(B) - P ( A Ç B)
2 ÷ø 32 32 \ P(B) = 3
32 10
3 ´ 10 By using P(B) = P(A È B) + P(A Ç B) – P(A)
32 32
(3) 5C 3 çæè 1 ö5 4 2 5 \ P( A Ç B ) = 2 - 1 = 5/12
2 ÷ø 4 ´ 5 3 4
32 32
5 C 4 çèæ 1 ö5 1 25. (a) : Given P(A) = 1/2 \ P ( A ) = 1/2
2 ÷ø 1 52
(4) 5 × 32 32 2
P(B) = 1/3 \ P (B ) = 3
5C 5 çèæ 1 ö5 1 3
2 ÷ø P(C) = \ P (C ) = 4
(5) 4
Now problem will be solved if any one of them will solve the
80 240 problem.
åf i = 1 å f i x i= 32 å f i x i 2 = 32 \ P(at least one of them solve the problem)
5 = 1 – probability none of them can solve the problem.
x = mean =
or P(A È B È C) = 1 – P( A) P(B) P(C )
2
åå åå Now variance = fi xi2 æ fi xi ö 2 = 1 – 1 . 2 . 3 = 3/4
ç ÷ 2 3 4
-
fi çè fi ÷ø
Trigonometry 105
CHAPTER TRIGONOMETRY
17
1. If x, y, z are in A.P. and tan– 1x , tan –1y and tan– 1z are also in If cos(b – g) + cos(g – a) + cos(a – b) = - 32 , then
(a) A is false and B is true
A.P., then (b) both A and B are true
(c) both A and B are false
(a) 2x = 3y = 6z (b) 6x = 3y = 2z (2013) (d) A is true and B is false
(c) 6x = 4y = 3z (d) x = y = z
2. ABCD is a trapezium such that AB and CD are parallel and (2009)
BC ^ CD. If ÐADB = q, BC = p and CD = q, then AB is equal to
p2 + q2 cos q p2 + q2 9. AB is a vertical pole with B at the ground level and A at the
(a) pcosq + qsinq (b) p2 cosq + q2 sinq top. A man finds that the angle of elevation of the point A
from a certain point C on the ground is 60°. He moves away
(p2 + q2 )sin q (p2 + q2 )sin q (2013) from the pole along the line BC to a point D such that
(c) (pcosq + qsinq )2 (d) pcosq + qsinq CD = 7 m. From D the angle of elevation of the point A is 45°.
Then the height of the pole is
3. The expres sion 1 tanA + 1 c o t A can be written as 73 1 m 73 1 m
- cotA - tan A (a) 2 3 + 1 (b) 2 3 - 1
(a) secA cosecA + 1 (b) tanA + cotA
(c) secA + cosecA(d) sinA cos A + 1 (2013) 7 3 7 3
(c) ( 3 + 1) m (d) ( 3 - 1) m (2008)
2 2
4. In a DPQR, if 3 sin P + 4 cos Q = 6 and 4 sin Q + 3cos P = 1, then
the angle R is equal to (c) 5p/6 (d) p/6 10. The value of is cot èæ cosec-1 5 + tan -1 32 öø
(a) p/4 (b) 3p/4 (2012) 3
5. If A = sin2 x + cos4 x , then for all real x 5 6 3 4
(a) 17 (b) 17 (c) 17 (d) 17
3 £ A £ 13 (2008)
4 16
(a) 1 £ A £ 2 (b) 11. A tower stands at the centre of a circular park. A and B are
two points on the boundary of the park such that AB (= a)
(c) 3 £ A £ 1 (d) 13 £ A £ 1 (2011) subtends an angle of 60° at the foot of the tower, and the
4 16 angle of elevation of the top of the tower from A or B is 30°.
The height of the tower is
6. Let cos(a + b) = 54 and let sin(a – b) = 153 , where 0 £ a , b £ p .
4
(a) a / 3 (b) a 3
Then tan2a =
25 56 19 20 (2010) (c) 2a / 3 (d) 2a 3. (2007)
16 33 12 17
(a) (b) (c) (d)
7. For a regular polygon, let r and R be the radii of the inscribed 12. The largest interval lying in çæè -p , p2 ÷øö for which the function,
2
and the circumscribed circles. A false statement among the f (x) = 4- x 2 + cos -1 æ 2x - 1øö + log (cos x) is defined, is
è
following is
(a) there is a regular polygon with r/R = 1/2 (a) êéë- p , p öø÷ (b) êéë0, p ÷øö
4 2 2
(b) there is a regular polygon with r/R = 1 / 2
(c) there is a regular polygon with r/R = 2 / 3 (c) [0, p] (d) çèæ - p , p ÷øö . (2007)
2 2
(d) there is a regular polygon with r/R = 3 / 2 (2010)
8. Let A and B denote the statements 13. If sin -1 èçæ x øö÷ + c osec - 1 çèæ 45 ø÷ö = p2 , then the values of x is
A : cos a + cos b + cos g = 0 5
B : sin a + sin b + sin g = 0 (a) 4 (b) 5 (c) 1 (d) 3. (2007)
106 JEE MAIN CHAPTERWISE EXPLORER
14. If 0 < x < p and cosx + sinx = 1/2, then tanx is 23. Let a, b be such that p < a – b < 3p. If
(1 - 7 ) (4 - 7) sina + sinb = –21/65, and cosa + cosb = –27/65, then the value
(a) (b)
of cos a - b is
4 3 2
(c) - (4 + 7) (1 + 7) (2006) (a) 6 (b) 3 (c) - 3 (d) -6 56 .
3 (d) . 65 130 130 (2004)
4
15. The number of value of x in the interval [0, 3p] satisfying the ( ) ( ) 24. If in a triangle ABC, acos2 C A = 32b , then the sides
equation 2sin2 x + 5sin x – 3 = 0 is 2 + c cos2 2
(a) 4 (b) 6 (c) 1 (d) 2. (2006)
16. If in a DABC, the altitudes from the vertices A, B, C on opposite a, b and c
sides are in H.P., then sinA, sinB, sinC are in (a) are in G.P. (b) are in H.P.
(a) H.P. (c) satisfy a + b = c (d) are in A.P. (2003)
(b) ArithmeticGeometric progression 25. In a triangle ABC, medians AD and BE are drawn. If AD = 4,
(c) A.P. ÐDAB = p/6 and ÐABE = p/3, then the area of the DABC is
(d) G.P. (2005) (a) 16/3 (b) 32/3
17. If cos-1 x - cos-1 y = a , then 4x2 – 4xy cosa + y2 is equal to (c) 64/3 (d) 8/3. (2003)
2 26. The upper 3/4t h portion of a vertical pole subtends an angle
(a) 4 (b) 2sin2a tan– 1( 3/5) at a point in the horizontal plane through its foot and
(c) –4sin2 a (d) 4sin2 a. (2005) at a distance 40 m from the foot. A possible height of the
= p . vertical pole is
2
18. In a triangle ABC, let ÐC If r is the inradius and R is (a) 40 m (b) 60 m
the circumradius of the triangle ABC, then 2(r + R) equals (c) 80 m (d) 20 m. (2003)
(a) a + b (b) b + c 27. The sum of the radii of inscribed and circumscribed circles for
an n sides regular polygon of side a, is
(c) c + a (d) a + b + c (2005)
( ) (a) ( ) (b)
( ) 19. p . P and tan çèæ Q2 ÷öø are a2 cot p a cot p
In a triangle PQR, if ÐR = 2 If tan 2 2 n 2 n
( ) (c) ( ) (d) p
the roots of ax2 + bx + c = 0, a ¹ 0 then a4 cot p a cot n . (2003)
2 n
(a) b = a + c (b) b = c
28. The trigonometric equation sin– 1x = 2sin– 1a , has a solution for
(c) c = a + b (d) a = b + c (2005)
1
20. A person standing on the bank of a river observes that the angle (a) all real values (b) | a | < 2
of elevation of the top of a tree on the opposite bank of the (c) | a | ³ 1 (d) 1 < | a | < 1 . (2003)
2 2 2
river is 60º and when he retires 40 meters away from the tree
the angle of elevation becomes 30º. The breadth of the river is 29. In a triangle with sides a, b, c, r 1 > r 2 > r 3 (which are the ex
radii) then
(a) 40 m (b) 30 m
(c) 20 m (d) 60 m. (2004) (a) a > b > c (b) a < b < c
21. The sides of a triangle are sina, cosa and 1+ sin a cosa for (c) a > b and b < c (d) a < b and b > c.
some 0 < a < p2 . Then the greatest angle of the triangle is (2002)
30. co t -1 éë (c o s a ) 1 ûù + ta n -1 éë ( c o s a 1 ûù = x
2
) 2
(a) 120º (b) 90º
(c) 60º (d) 150º. (2004) then sin x = (b) cot2 (a/2)
(a) 1 (d) cot(a/2)
22. If u = a 2 cos 2 q + b 2 sin 2 q + a 2 sin 2 q + b 2 cos 2 q (c) tana (2002)
then the difference between the maximum and minimum values 31. The number of solutions of tan x + sec x = 2cos x in [0, 2p)
of u 2 is given by is
(a) (a + b) 2 (b) 2 a 2 + b 2 (a) 2 (b) 3 (c) 0 (d) 1.
(c) 2(a 2 + b 2) (d) (a – b)2 . (2004) (2002)
Answer Key
1. (d) 2. (d) 3. (a) 4. (d) 5. (c) 6. (b)
7. (c) 8. (b) 11. (a) 12. (b)
13. (d) 14. (c) 9. (c) 10. (b) 17. (d) 18. (a)
19. (c) 20. (c) 23. (c) 24. (d)
25. (*) 26. (a) 15. (a) 16. (c) 29. (a) 30. (a)
31. (b)
21. (a) 22. (d)
27. (a) 28. (*)
Trigonometry 107
1. (d) : As x, y, z are in A.P. Þ 2y = x + z … (i) 5. (c) : A = sin 2 x + cos 2 x
tan– 1x , tan –1 y and tan –1z are in A.P., then We have cos 4 x £ cos 2 x
2tan –1 y = tan –1 x + tan –1z sin 2 x = sin 2 x
Adding sin 2x + cos 4 x £ sin 2 x + cos 2 x
2 tan -1 y = tan -1 èæç x + z ö \ A £ 1.
1 - xz ÷ø Again A = t + (1 – t) 2 = t 2 – t + 1, t ³ 0,
where minimum is 3/4
Þ tan -1 æ 2 y ö = tan -1 æçè x + z ö
çè 1 - y 2 ÷ø 1 - xz ø÷
Thus y 2 = xz … (ii)
From (i) and (ii), we get x = y = z. Thus 3 £ A £ 1 .
4
Remark : y ¹ 0 is implicit to make any of the choice correct.
6. (b) : cos(a + b) = 4/5 giving tan(a + b) = 3/4
2. (d) : Using sine rule in triangle ABD, we get
Also sin(a – b) = 5/13 given tan(a – b) = 5/12
p2 + q2 sinq
AB = BD Þ AB = 3 + 5
4 12
sinq sin(q + b) sin(q + b) = tan(a + b) + tan(a - b) = = 36 + 20 = 56
As tanb = p , we have 1 - tan(a + b)tan(a - b) 1 - 3 ´ 5 48 - 15 33
q 4 12
sin(q + b) = sinq cosb + cosq sinb 7. r = cos p ,
(c) : We have R n
= sinq × q + cosq × p = pcosq + qsinq Let cos p = 1 .
p2 + q2 p2 + q2 p2 + q2 n2
We then get AB = (p2 + q2 )sin q Thus we get p = p
pcosq + qsinq n4
i.e., n = 4, acceptable.
(a) : tanA + c o t A cos p = 1 Þ p = p . \ n = 3, acceptable.
3. - cotA - tan A n2 n3
1 1
= sin2 A + cos 2 A cos p = 3 Þ p = p . \ n = 6, acceptable.
cosA(sinA - cosA) sinA(cosA - sinA) n2 n6
= sin3 A - cos 3 A But cos p = 2 will produce no value of n.
(sinA - cosA)cosAsinA n3
= (sinA - cosA)(sin2 A + sinAcosA + cos2 A) As 1 < 2 < 1 Þ cos p < cos p < cos p
(sinA - cosA)sinAcosA 23 2 3 n 4
= 1 + sinAcos A = 1 + secAcosec A Þ p > p > p Þ 3 < n < 4 (impossible)
sinAcosA 3n4
4. (d) :3 sin P + 4cos Q = 6 8. (b) : cos(b – g) + cos(g – a) + cos(a – b) = – 3/2
Þ (cosb cosg + sinb sing) + (cosg cosa + sing sina)
4 sin Q + 3cos P = 1 + (cosa cosb + sina sinb) = – 3/2
Þ 2(cosb cosg + cosg cosa + cosa cosb)
Þ 16 + 9 + 24 (sin (P + Q)) = 37 + 2( sinb sing + sing sina + sina sinb) + 3 = 0
Þ {cos 2a + cos 2b + cos 2g
Þ 24 (sin (P + Q)) = 12 + 2(cosa cosb + cosb cosg + cosg cosa)}
+ {sin 2a + sin 2b + sin 2g
Þ sin (P + Q) = 1 Þ sin R = 1 Þ R = 5 p or p + 2(sina sinb + sinb sing + sing sina)} = 0
2 2 66 Þ (cosa + cosb + cosg)2 + (sina + sinb + sing)2 = 0
Which yields simultaneously
But if R = 5 p then P < p and then 3 sin P < 1 cosa + cosb + cosg = 0 and sina + sinb + sing = 0
6 6 2
and so 3sin P + 4 cos Q < 1 + 4 (¹ 6) Thus, R = p .
2 6
108 JEE MAIN CHAPTERWISE EXPLORER
9. (c) : 1 st Solution : 13. (d) : sin -1 èçæ x öø÷ + sin -1 çæè 54 ÷öø = p
Let height of the pole AB be h. Then 5 2
BC = h cot 60° = h / 3
BD = h cot 45° = h Þ sin -1 çèæ x ÷øö = p - sin -1 çèæ 54 ÷öø
As BD – BC = CD 5 2
A x = sin æ p - sin -1 çæè 54 ÷öø ö
5 çè 2 ÷ø
x = cosèçæ sin-1 4 ÷øö = cos èçæ cos -1 3 ÷öø = 3 Þ x = 3.
5 5 5 5
h 14. (c) : 0 < x < p
B 60° 45° Given cos x + sin x = 1
C 7 D 2
Þ h- h = 7 Þ h( 3 -1) = 7 3 Þ 1 + sin 2 x = 1 (By squaring both sides)
3 4
2 tan x = -3
1 + tan 2 x 4
7 3 =7 3( 3 + 1) 7 2 3 (
Þ h= 3 - 1 2 = 3 + 1)m
Þ 3tan2 x + 8tan x + 3 = 0
2n d Solution : tan x = -8 ± 64 - 36 = -4 ± 7
We use the fact that the ratio of distance of B from D and that 6 3
of B from C i.e. BD to BC is 3 :1
Q tan x < 0 Þ tan x = -4 - 7
8
BD = 3, so that BD = 3
BC CD 3 - 1 15. (a) : 2sin2 x + 5 sinx –3 = 0
A Þ sin x = 12 ,sin x ¹ - 3
there sin x = 21 , we know that each trigonometrical function
B 60° 45° assumes same value twice in 0 £ x £ 360°.
C D In our problem 0° £ x £ 540°. So number of values are 4
7 like 30°, 150°, 390°, 510°.
Then BD = 3 CD = 3 × 7 = 7 3 ( 3 + 1)
3 -1 3 - 1 2
16. (c) : Altitude from A to BC is AD
A
1 AD ´ BC
As AB = BD, the height of the pole = 7 3 ( 3 + 1)m Area of D = 2
2
\ 2 × Area of D = AD
10. (b) : cot æè cosec -1 5 + tan -1 2 ö a
3 3 ø
\ Altitudes are in H.P.
= cot èæ tan -1 3 + tan -1 2 ö \ 2D , 2D , 2 D Î HP B D C
4 3 ø abc a
æ 3 + 2 ö Þ 1 , 1 , 1 Î H.P. Þ a, b, c Î A.P.
4 3 ÷ a b c
ç tan -1 tan -1 167 öø 6
= cot ç 3 2 ÷ = cot æè = 17 17. (d) : Using cos-1 A - cos -1 B
çè 1 - 4 × 3 ÷ø ( ) = cos-1 AB + (1 - A2 ) (1 - B2 )
11. (a) : OP = Tower
OAB is equilateral triangle \ cos-1 x - cos -1 y = a
2
\ OA = OB = AB = a
In D AOP, A 30°a O Þ xy + 1 - x2 1 - y 2 = cos a
tan 30° = OOPA 2 4
60°
a a Þ çæè cos a - xy 2 = (1 - x 2 ) æ y 2 ö
2 çè1 - 4 ÷ø
Þ OP = a 30° ÷öø
3 B
Þ 4x 2 – 4xy cosa + y 2 = 4 (1 – cos 2a) = 4sin 2a.
12. (b) : f (x) is defined if -1 £ 2x -1 £ 1 and cos x > 0
18. (a) : c = 2 R
sin C
p p \ 0 £ x < p2 .
or 0 £ x £ 4 and - 2 < x < 2 \ c = 2R ...(A)(Q C = 90º) and
Trigonometry 109
tan C = r A (21)2 + (27) 2
2 s - c 2(1 + cos a cos b + sin a sin b) = (65)2
\ r = (s – c) r 1170
2[1 + cos (a – b)] = (65) 2
R
( ) tanC B cos 2 (a - b ) 1170 130 ´ 9 9
2 = tan 45° = 1 2 = 4 ´ 65 ´ 65 = (130) ´ (130) =
...(B)
C 130
= a + b + c - c ; 2r = a + b – c a - b 3
2 \ cos =
2 130
adding (A) and (B) we get 2(r + R) = a + b.
19. (c) : ÐR = 90° \ ÐP + ÐQ = 90° As p < a – b < 3p then cos æ a - b ÷öø = negative
çè 2
P 920 - Q2 , P Q
\ 2 = 2 = 45 - 2 C
24. (d) : 2a cos2 + 2c cos2 A /2 =3b
tan P / 2 = 1 - tanQ / 2 (from given)
1 1 + tan Q / 2 2
Þ
Þ a(1 + cos C) + c(1 + cos A) = 3b
tan P + tan Q = 1 - tan P × tan Q Þ a + c + a cos C + c cos A = 3b
Þ 22 2 2 (a cos C + c cos A = b projection formula)
Þ a + c + b = 3b
Þ - b = 1 - c Þ a + c = 2b A
aa
çèæ Q tan P2 , tan Q2 are roots of ax2 + bx + c = 0 ÷öø 25. (*) : OB = tan 30° E
AO
p/3 O
c - b A uuur OA = 8 3
Þ a = 1 Þ c = a + b. 30° Þ OB = 3 9 B D C
20. (c) : Breadth of river OC = AC cos 60° Area of triangle ADB 1 ´ 8 3 ´ 4 = 16 3
=
29 9
30° 60°
= 40 cos 60° B C O 16 3 = 32 3
Area of triangle ABC = 2 ×
21. (a) : If a2 = sin2 a, b2 = cos2 a, c 2 = 1 + sin a cos a 99
- sin a cos a 26. (a) : a = A + b 3/4 h
then cos c = 2 sin a cos a \ cos c = –1/2 \ b = A – a b 1/4 h
tan A - tan a aA
22. (d) : u 2 = a2 ( cos 2q + sin2 q) + b2 ( sin2 q + cos2 q) + tan b = 1 - tan A tan a
( ) ( ) 2 a4 + b4 sin 2 q cos2 q + a 2b2 sin 4 q + cos 4 q 3 h + æ - h ö
5 40 çè 160 ÷ø
Þ =
2 æ h ö æ hö
( ) = a 2 + b 2 + 2 2 çè 40 ÷ø çè 160 ÷ø
a2b2 + a2 - b2 sin q co s 2 q 1 - -
= a2 + b2 + 2 + a 2b2 + æ a2 - b 2 ö 2 q Þ h2 – 200h + 6400 = 0
èçç 2 ø÷÷ sin Þ (h – 40) (h – 160) = 0
Þ h = 40 or h = 160
\ u2 will be maximum or minimum according as q = p/4 or
q = 0° 27. (a) : If R be the radius of circumcircle of regular polygon of
n sides, and r be the radius of inscribed circle then
\ Max. u 2 = 2(a2 + b2 ) and
Min. u 2 = a 2 + b2 + 2ab = (a + b)2 a p a p
Now Maximum u2 – Minimum u2 R = 2 cosec 2n and r = 2 cot n
= 2(a 2 + b 2 ) – (a2 + b2 + 2ab) a æ p p ö
2 çè n cot ÷ø
= a 2 + b 2 – 2ab = (a – b) 2 \ R + r = cosec +
n
23. (c) : sin a + sin b = - 21
65 æ 1 +sicno pns pn ÷ö÷÷ø÷ A O B
ç
and cos a + cos b = - 27 = a ç = a cot p R π π R OL = r
65 2 ççè 2 2 n n n OB = R
by squaring and adding we get B L C
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MCQ IIT JEE
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